6. Directional Derivatives and Gradients

Exercises

a. Derivative along a Curve and the Gradient

Find the derivative of the given function along the given curve at the given time.

\(f(x,y)=2x^2y^3\), \(\vec r(t)=(t^3,t^2)\) and \(t=2\). Use two methods.
1. Use the gradient formula.
  
  The derivative along a curve is the dot product of the velocity of the curve and the gradient of the function.
  
  \(\dfrac{df}{dt}(2)=3\cdot2^{14}\)
  
  The derivative is the dot product of the velocity of the curve and the gradient of the function. The velocity is: \[\begin{aligned} \vec v(t)&=\langle3t^2,2t\rangle \\ \vec v(2)&=\langle12,4\rangle \\ \end{aligned}\] The gradient is \[\begin{aligned} \vec\nabla f &=\langle 4xy^3, 6x^2y^2\rangle \\ \left.\vec\nabla f\right|_{\vec r(t)}&=\langle 4t^9, 6t^{10}\rangle\\ \left.\vec\nabla f\right|_{\vec r(2)}&=\langle 2^{11}, 3\cdot2^{11}\rangle \end{aligned}\] Their dot product is: \[\begin{aligned} \dfrac{df}{dt}(2) &=\vec v(2)\cdot\left.\vec\nabla f\right|_{\vec r(2)} =\langle12,4\rangle\cdot\langle 2^{11}, 3\cdot2^{11}\rangle \\ &=3\cdot2^{13}+3\cdot2^{13}=3\cdot2^{14} \end{aligned}\]
  
  as
2. Differentiate the composition.
  
  \(\dfrac{df}{dt}(2)=3\cdot2^{14}\)
  
  as
  
  Since \(x\) and \(y\) are known in terms of \(t\), we substitute \(x=t^3\) and \(y=t^2\) into the function \(f(x,y)=2x^2y^3\): \[ f(\vec r(t)=2(t^3)^2(t^2)^3=2t^{12} \] Then the derivative is: \[ \dfrac{df}{dt}=24t^{11} \] When \(t=2\), this becomes: \[ \dfrac{df}{dt}(2)=24\cdot2^{11}=3\cdot2^{14} \]
  
  as
\(f(x,y,z)=2x^2y^3z\), \(r(t)=(t^3,t^2,t)\) and \(t=1\). Use two methods.
1. Use the gradient formula.
  
  \(\dfrac{df}{dt}=26\)
  
  as
  
  The derivative is the dot product of the velocity of the curve and the gradient of the function. The velocity is: \[\begin{aligned} \vec v(t)&=(3t^2,2t,1) \\ \vec v(1)&=(3,2,1) \end{aligned}\] The gradient is: \[\begin{aligned} \vec\nabla f&=\langle 4xy^3 z,6x^2y^2 z,2x^2y^3\rangle \\ \left.\vec\nabla f\right|_{\vec r(t)}&=\langle 4t^{10}, 6t^{11}, 2t^{12}\rangle \\ \left.\vec\nabla f\right|_{\vec r(1)}&=\langle 4,6,2\rangle \end{aligned}\] Their dot product is: \[\begin{aligned} \dfrac{df}{dt}(1)&=\vec v(1)\cdot\left.\vec\nabla f\right|_{\vec r(2)} =\langle 3, 2, 1\rangle\cdot\langle 4,6,2\rangle \\ &=3\cdot 4+2\cdot 6+1\cdot 2=26 \end{aligned}\]
  
  jp
2. Differentiate the composition.
  
  \(\dfrac{df}{df}(1)=26\)
  
  jp
  
  Since \(x\), \(y\), and \(z\) are already known in terms of \(t\), we can substitute \(x=t^3,\quad y=t^2,\quad z=t\) into the function \(f(x,y,z)=2x^2y^3z\) \[ f(\vec r(t))=2(t^3)^2(t^2)^3(t)=2t^{13} \] Then the derivative is: \[ \dfrac{df}{dt}=26t^{12} \] When \(t=1\), this becomes: \[ \dfrac{df}{dt}(1)=26(1)^{12}=26 \]
  
  jp

Find the derivative of the given function along a curve at the instant when it passes through the given position with the given velocity.

\(f(x,y)=\sin(x)\cos(y)\), \(\vec r(2)=\left(\dfrac{\pi}{3},\dfrac{\pi}{6}\right)\) and \(\vec v(2)=\langle 1,2\rangle\).

\(\dfrac{df}{dt}=-\,\dfrac{\sqrt{3}}{4}\)

jp

The derivative is the dot product of the velocity and the gradient at the point. We compute the gradient and evaluate it at the point: \[\begin{aligned} \nabla f &=\langle \cos(x)\cos(y),-\sin(x)\sin(y)\rangle \\ \left.\nabla f\right|_{(\pi/3,\pi/6)} &=\left\langle \dfrac{\sqrt{3}}{4},-\,\dfrac{\sqrt{3}}{4}\right\rangle \end{aligned}\] Its dot product with the velocity gives the derivative: \[\begin{aligned} \dfrac{df}{dt} &=\vec v(2)\cdot\left.\nabla f\right|_{(\pi/3,\pi/6)} \\ &=\dfrac{\sqrt{3}}{4}-2\dfrac{\sqrt{3}}{4} =-\,\dfrac{\sqrt{3}}{4} \end{aligned}\]

jp
\(f(x,y,z)=z^2 e^{xy}\), \(\vec r(1)=(3,2,1)\) and \(\vec v(1)=\langle 1,2,3\rangle\).

\(\dfrac{df}{dt}(1)=14e^6\)

as

The derivative is the dot product of the velocity and the gradient at the point. We compute the gradient and evaluate it at the point: \[\begin{aligned} \nabla f =&\langle yz^2e^{xy},xz^2e^{xy},2ze^{xy}\rangle \\ \left.\nabla f\right|_{(3,2,1)} &=\langle 2e^6,3e^6,2e^6\rangle\\ \end{aligned}\] Its dot product with the velocity gives the derivative: \[\begin{aligned} \dfrac{df}{dt}(1) &=\vec v(1)\cdot \left.\nabla f\right|_{(3,2,1)} \\ &=\langle 1,2,3\rangle\cdot\langle 2e^6,3e^6,2e^6\rangle \\ &=2e^6+6e^6+6e^6 =14e^6 \end{aligned}\]

as

A ant is crawling on a frying pan where the temperature is \(T=300+\sqrt{x^2+y^2}\). If the ant is at \(P(x,y)=(3,-2)\) and has velocity \(\vec v=\langle 2,-3\rangle\), find the rate of change of the temperature as seen by the ant.

\(\dfrac{dT}{dt}=\dfrac{12}{\sqrt{13}}\)

The gradient of the temperature is \[\begin{aligned} \vec\nabla T=\left\langle \dfrac{\partial T}{\partial x},\dfrac{\partial T}{\partial y}\right\rangle &=\left\langle\dfrac{x}{\sqrt{x^2+y^2}},\dfrac{y}{\sqrt{x^2+y^2}}\right\rangle \\ \left.\vec\nabla T\right|_{(3,-2)} &=\left\langle\dfrac{3}{\sqrt{13}},-\,\dfrac{2}{\sqrt{13}}\right\rangle \end{aligned}\] We compute the dot product with \(\vec v=\langle 2,-3\rangle\) to find \[\begin{aligned} \dfrac{dT}{dt} &=\vec v\cdot\left.\vec\nabla T\right|_{(3,-2)} =\dfrac{6}{\sqrt{13}}+\dfrac{6}{\sqrt{13}} =\dfrac{12}{\sqrt{13}} \end{aligned}\]

as,jp
Ham Duet is flying the Centurion Eagle through a dark matter field whose density is \(\delta=\dfrac{x^2y^3}{z}\). Find the rate of change of the density if the Eagle is located at \((3,2,1)\) and has velocity \(\vec v=\langle 0.3,0.4,0.2\rangle\).

\(\dfrac{d\delta}{dt}=43.2\)

as

The gradient of the density is \[\begin{aligned} \vec\nabla\delta=\left\langle\delta_x,\delta_y,\delta_z\right\rangle &=\left\langle\dfrac{2xy^3}{z},\dfrac{3x^2y^2}{z},-\,\dfrac{x^2y^3}{z^2}\right\rangle \\ \left.\vec\nabla\delta\right|_{(3,2,1)} &=\langle48,108,-72\rangle \end{aligned}\] Then the rate of change of the density is the derivative along the velocity: \[\begin{aligned} \dfrac{d\delta}{dt}&=\vec v\cdot\left.\vec\nabla\delta\right|_{(3,2,1)} \\ &=\langle 0.3,0.4,0.2\rangle\cdot\langle48,108,-72\rangle \\ &=14.4+43.2-14.4=43.2 \end{aligned}\]

as, jp
A weather balloon is located at \(Q=(x,y,z)=(20,-35,15)\,\text{m}\) and has velocity \(\vec v=\langle 6,4,3\rangle\,\dfrac{\text{m}}{\text{hr}}\). It measures the pressure to be \(P|_Q=1.2\,\text{atm}\) and it gradient to be \(\vec\nabla P|_Q=\langle.02,.03,-.05\rangle\,\dfrac{\text{atm}}{\text{m}}\). Find the rate of change of the pressure as seen by the balloon at \(Q\).

\(\dfrac{dP}{dt}==.09\,\dfrac{\text{atm}}{\text{sec}}\)

We know the gradient at \(Q\) is \(\vec\nabla P|_Q=\langle.02,.03,-.05\rangle\) and the velocity is \(\vec v=\langle 6,4,3\rangle\). So the rate of change of the pressure is \[\begin{aligned} \left.\dfrac{dP}{dt}\right|_Q &=\vec v\cdot\left.\vec\nabla P\right|_Q \\ &=\langle 6,4,3\rangle\cdot\langle.02,.03,-.05\rangle \\ &=.12+.12-.15=.09\,\dfrac{\text{atm}}{\text{sec}} \end{aligned}\]
A first airplane flies along the curve \(\vec r_1(t)=\langle t,t^2,t^3\rangle\). At time \(t=1\,\text{hr}\), the pilot of the first plane measures the rate of change of the temperature to be \(\left.\dfrac{d(T\circ\vec r_1)}{dt}\right|_{t=1} =12\,\dfrac{^\circ\text{C}}{\text{hr}}\).

A second airplane flies along the curve \(\vec r_2(t) =\left\langle \dfrac{t}{2},\dfrac{t^2}{4},\dfrac{t^3}{8}\right\rangle\). Find the rate of change of the temperature as seen by the second pilot at time \(t=2\,\text{hr}\), that is: \(\left.\dfrac{d(T\circ\vec r_2)}{dt}\right|_{t=2}\).

(Assume that the temperature depends on position \((x,y,z)\) but not time \(t\).)

Notice that both planes make their measurments at the same point. How are their velocities related?

\(\left.\dfrac{d(T\circ\vec r_2)}{dt}\right|_{t=2} =6\,\dfrac{^\circ\text{C}}{\text{hr}}\)

Here are the two curves and their velocities: \[\begin{aligned} \vec r_1(t)&=\langle t,t^2,t^3\rangle &\qquad\vec v_1(t)&=\langle 1,2t,3t^2\rangle\\ \vec r_2(t)&=\left\langle \dfrac{t}{2},\dfrac{t^2}{4},\dfrac{t^3}{8}\right\rangle &\qquad\vec v_2(t)&=\left\langle \dfrac{1}{2},\dfrac{2t}{4},\dfrac{3t^2}{8}\right\rangle\\ \end{aligned}\] At the appropriate times, these are: \[\begin{aligned} \vec r_1(1)&=\langle 1,1,1\rangle &\qquad\vec v_1(1)&=\langle 1,2,3\rangle\\ \vec r_2(2)&=\langle 1,1,1\rangle &\qquad\vec v_2(2)&=\langle \dfrac{1}{2},\dfrac{2}{2},\dfrac{3}{2}\rangle =\dfrac{1}{2}\langle 1,2,3\rangle \end{aligned}\] Notice the points are the same and the second velocity is just a multiple of the first: \[ \vec v_2(2)=\dfrac{1}{2}v_1(1) \] So the derivatives along the two curves are related by: \[\begin{aligned} \left.\dfrac{d(T\circ\vec r_2)}{dt}\right|_{t=2} &=\vec v_2(2)\cdot\left.\vec\nabla T\right|_{(1,1,1)} =\dfrac{1}{2}v_1(1)\cdot\left.\vec\nabla T\right|_{(1,1,1)} \\ &=\dfrac{1}{2}\left.\dfrac{d(T\circ\vec r_1)}{dt}\right|_{t=1} =\dfrac{1}{2}\cdot12=6\,\dfrac{^\circ\text{C}}{\text{hr}} \end{aligned}\]

b. Derivative along a Vector

Find the derivative of the given function along the given vector at the given point.

\(f(x,y)=\dfrac{y}{x^2}\), \(\vec v=\langle 3,4\rangle\) and \(P=(2,3)\).

\(\dfrac{df}{dt}=-\,\dfrac{5}{4}\)

We first calculate the partial derivatives: \[ \dfrac{\partial f}{\partial x}=-\,\dfrac{2y}{x^3} \] and \[ \dfrac{\partial f}{\partial y}=\dfrac{1}{x^2} \] We combine them into the gradient: \[\begin{aligned} \vec\nabla f =\left\langle\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right\rangle &=\left\langle -\,\dfrac{2y}{x^3},\dfrac{1}{x^2}\right\rangle \end{aligned}\] At point \(P=(2,3)\) this is: \[\begin{aligned} \left.\vec\nabla f\right|_{(2,3)} &=\left\langle -\,\dfrac{3}{4},\dfrac{1}{4}\right\rangle \end{aligned}\] With \(\vec v=\langle 3,4\rangle\) this becomes: \[\begin{aligned} \left .\vec\nabla_{\vec v}f\right|_P =\left .\vec v\cdot\vec\nabla f\right|_P &=\langle 3,4\rangle\cdot\left\langle -\,\dfrac{3}{4},\dfrac{1}{4}\right\rangle \\ &=-\,\dfrac{9}{4}+1=-\,\dfrac{5}{4} \end{aligned}\]

as/jp
\(f(x,y,z)=x\sin(yz)-y\cos(xz)\), \(\vec v=\langle 2,2,9\pi\rangle\) and \(P=\left(\dfrac{1}{3},\dfrac{1}{6},\pi\right)\).

\(\nabla_{\vec v} f=\sqrt{3}\pi\)

The partial derivatives of \(f\) are: \[\begin{aligned} &\dfrac{\partial f}{\partial x}=\sin(yz)+yz\sin(xz) \\[4pt] &\dfrac{\partial f}{\partial y}=xz\cos(yz)-\cos(xz) \\[2pt] &\dfrac{\partial f}{\partial z}=xy\cos(yz)+xy\sin(xz) \end{aligned}\] At point \(P=\left(\dfrac{1}{3},\dfrac{1}{6},\pi\right)\), the gradient is: \[\begin{aligned} \left.\vec\nabla f\right|_{(1/3,1/6,\pi)} &=\left\langle\dfrac{1}{2}+\dfrac{\sqrt{3}\pi}{12},\dfrac{\sqrt{3}\pi}{6}-\dfrac{1}{2},\dfrac{\sqrt{3}}{18}\right\rangle \end{aligned}\] So the derivative along \(\vec v=\langle 2,2,9\pi\rangle\) is: \[\begin{aligned} \nabla_{\vec v} f =\vec v\cdot\vec\nabla f &=1+\dfrac{\sqrt{3}\pi}{6}+\dfrac{\sqrt{3}\pi}{3}-1+\dfrac{\sqrt{3}\pi}{2}\\ &=\sqrt{3}\pi \end{aligned}\]

as/jp
\(k(x,y)=\dfrac{x^2+3y}{(x+3)^2}\), \(\vec v=\left\langle 10,23\right\rangle\) and \(P=(3,2)\)

\(\dfrac{dk}{dt}=\dfrac{79}{36}\)

The partial derivatives of \(k\) are: \[\begin{aligned} \dfrac{\partial k}{\partial x} &=\dfrac{(x+3)^22x-(x^2+3y)2(x+3)}{(x+3)^4} \\[2pt] &=\dfrac{(x+3)2x-(x^2+3y)2}{(x+3)^3} \\[2pt] &=\dfrac{2x^2+6x-2x^2-6y}{(x+3)^3} =\dfrac{6x-6y}{(x+3)^3} \\[5pt] \dfrac{\partial k}{\partial y} &=\dfrac{(x+3)^23-(x^2+3y)0}{(x+3)^4} =\dfrac{3}{(x+3)^2} \end{aligned}\] So the gradient of \(k\) is: \[\begin{aligned} \vec\nabla k =\left\langle\dfrac{\partial k}{\partial x},\dfrac{\partial k}{\partial y} \right\rangle &=\left\langle\dfrac{6x-6y}{(x+3)^3},\dfrac{3}{(x+3)^2}\right\rangle \end{aligned}\] At the point \(P=(3,2)\) this is \[ \left.\vec\nabla k\right|_P =\left\langle\dfrac{6}{6^3},\dfrac{3}{6^2}\right\rangle =\left\langle\dfrac{1}{36},\dfrac{1}{12}\right\rangle \] So the derivative of \(k\) along \(\vec v\) is: \[\begin{aligned} \left .\vec\nabla_{\vec v}k\right|_P =\left .\vec v\cdot\vec\nabla k\right|_P &=\langle 10,23\rangle\cdot\left\langle\dfrac{1}{36},\dfrac{1}{12}\right\rangle \\ &=\dfrac{10}{36}+\dfrac{23}{12}=\dfrac{79}{36} \end{aligned}\]

jp
\(f(x,y)=\dfrac{3x-y}{3x^2-3y^2}\), \(\vec v=\langle3,6\rangle\) and \(P=(2,1)\)

\(\dfrac{df}{dt}=\dfrac{1}{3}\)

We first must calculate the partial derivatives of \(f\): \[\begin{aligned} \dfrac{\partial f}{\partial x} &=\dfrac{(3x^2-3y^2)(3)-(3x-y)(6x)}{(3x^2-3y^2)^2} \\ &=\dfrac{(3x^2-3y^2)-(3x-y)(2x)}{3(x^2-y^2)^2} \\ &=\dfrac{-3x^2+2xy-3y^2}{3(x^2-y^2)^2} \end{aligned}\] and \[\begin{aligned} \dfrac{\partial f}{\partial y} &=\dfrac{(3x^2-3y^2)(-1)-(3x-y)(-6y)}{(3x^2-3y^2)^2} \\ &=\dfrac{-(x^2-y^2)+(3x-y)(2y))}{3(x^2-y^2)^2} \\ &=\dfrac{-x^2+6xy-y^2}{3(x^2-y^2)^2} \\ \end{aligned}\] so the gradient of \(f\) is: \[\begin{aligned} \vec\nabla f =\left\langle\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y} \right\rangle &=\left\langle\dfrac{-3x^2+2xy-3y^2}{3(x^2-y^2)^2}, \dfrac{-x^2+6xy-y^2}{3(x^2-y^2)^2}\right\rangle \end{aligned}\] At point \(P=(2,1)\) this is: \[\begin{aligned} \left.\vec\nabla f\right|_P =\left\langle\dfrac{-12+4-3}{27},\dfrac{-4+12-1}{27} \right\rangle =\left\langle -\,\dfrac{11}{27},\dfrac{7}{27}\right\rangle \end{aligned}\] So the derivative of \(f\) along \(v\) is: \[\begin{aligned} \left .\vec\nabla_{\vec v}f\right|_P =\left .\vec v\cdot\vec\nabla f\right|_P &=\langle 3,6\rangle\cdot\left\langle -\,\dfrac{11}{27},\dfrac{7}{27}\right\rangle \\ &=\dfrac{-11+14}{9}=\dfrac{1}{3} \end{aligned}\]

jp
\(f(x,y,z)=x^2+y^2+z^2-x-y-z\), \(\vec v=\langle 3,2,1\rangle\), and \(P=(1,-3,-4)\)

\(\left .\nabla_{\vec v}f\right|_P=-20\)

We compute the gradient and evaluate it at the point \(P=(1,-3,-4)\): \[\begin{aligned} \nabla f &=\langle \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\rangle \\ &=\langle 2x-1,2y-1,2z-1\rangle \\ \left.\nabla f\right|_{(1,-3,-4)} &= \langle 2(1)-1, 2(-3)-1, 2(-4)-1\rangle \\ &=\langle 1,-7,-9\rangle \end{aligned}\] So the derivative along \(\vec v\) is: \[\begin{aligned} \left.\nabla_{\vec v}f\right|_P &=\left.\vec v\cdot\vec\nabla f\right|_P =\langle 3,2,1\rangle\cdot\langle 1,-7,-9\rangle \\ &=3-14-9=-20 \end{aligned}\]

c. Directional Derivatives

Find the directional derivative of the given function in the direction of the given vector at the given point.

\(f(x,y)=\dfrac{y}{x^2}\), \(\vec v=\langle 3,4\rangle\) and \(P=(2,3)\).

\(\nabla_{\hat v} f=-\,\dfrac{1}{4}\)

We first compute the gradient and evaluate it at \(P\): \[\begin{aligned} \vec\nabla f =\langle f_x,f_y\rangle &=\left\langle -\,\dfrac{2y}{x^3},\dfrac{1}{x^2}\right\rangle \\ \left.\vec\nabla f\right|_{(2,3)} &=\left\langle -\,\dfrac{3}{4},\dfrac{1}{4}\right\rangle \end{aligned}\] The directional derivative is the derivative using the unit vector. Since \(\vec v=\langle3,4\rangle\), the unit vector is: \[ \hat v =\dfrac{\vec v}{|\vec v|} =\left\langle \dfrac{3}{5},\dfrac{4}{5}\right\rangle \] So the directional derivative is: \[ \nabla_{\hat v} f =\hat v\cdot\vec\nabla f =-\,\dfrac{9}{20}+\dfrac{4}{20} =-\,\dfrac{1}{4} \]

as

Notice that this problem has the same function, vector and point as a previous problem, but this time we computed the directional derivaitve instead of the derivative along the whole vector including its magnitude. Since the magnitude of \(\vec v\) is \(5\), the answer to the previous problem is \(5\) times the answer to this problem.
\(f(x,y,z)=(2x+y)z^2\), \(\vec v=\langle 1,2,-2\rangle\) and \(P=(2,4,3)\).

\(\nabla_{\hat v} f=-20\)

as/jp

The gradient and its value at \(P\) are: \[\begin{aligned} \vec\nabla f =\langle f_x,f_y,f_z\rangle &=\left\langle 2z^2,z^2,2(2x+y)z\right\rangle \\ \left.\vec\nabla f\right|_{(2,4,3)} &=\left\langle 18,9,48\right\rangle \end{aligned}\] To find the directional derivative, we use the unit vector direction: \[ |\vec v|=\sqrt{1+4+4}=3 \] \[ \hat v =\dfrac{\vec v}{|\vec v|} =\left\langle \dfrac{1}{3},\dfrac{2}{3},-\,\dfrac{2}{3}\right\rangle \] So the directional derivative is: \[\begin{aligned} \nabla_{\hat v} f &=\hat v\cdot\vec\nabla f =6+6-32\\ &=-20 \end{aligned}\]

as/jp
\(f(x,y)=\dfrac{e^2x^2}{y^3}\), \(\vec v=\langle 2,1\rangle\), and \(P=(3,-2)\)

\(\left .\vec\nabla_{\hat v}f\right|_P=-\,\dfrac{51e^2}{16\sqrt{5}}\)

We first calculate the gradient of \(f\): \[\begin{aligned} \vec\nabla f =\left\langle\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y} \right\rangle &=\left\langle\dfrac{2e^2x}{y^3},-\,\dfrac{3e^2x^2}{y^4}\right\rangle \end{aligned}\] At point \(P=(3,-2)\) this is: \[\begin{aligned} \left.\vec\nabla f\right|_P =\left\langle-\,\dfrac{3e^2}{4},-\,\dfrac{27e^2}{16}\right\rangle \end{aligned}\] The directional derivative uses the unit vector: \[ \hat v =\dfrac{\vec v}{|\vec v|} =\left\langle \dfrac{2}{\sqrt{5}},\dfrac{1}{\sqrt{5}}\right\rangle \] So the directional derivative is: \[\begin{aligned} \left .\vec\nabla_{\hat v}f\right|_P &=\left .\hat v\cdot\vec\nabla f\right|_P \\ &=\left\langle \dfrac{2}{\sqrt{5}},\dfrac{1}{\sqrt{5}}\right\rangle \cdot\left\langle-\,\dfrac{3e^2}{4},-\,\dfrac{27e^2}{16}\right\rangle \\ &=\dfrac{-24e^2-27e^2}{16\sqrt{5}}=-\,\dfrac{51e^2}{16\sqrt{5}} \end{aligned}\]
\(f(x,y)=x^2-y^2-x+y\), \(\vec v=\left\langle \dfrac{2}{3},\dfrac{1}{3} \right\rangle\), and \(P=(3,8)\)

\(\left.\nabla_{\hat v}f\right|_P=-\sqrt{5}\)

We first calculate the gradient of \(f\): \[\begin{aligned} \vec\nabla f =\left\langle\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y} \right\rangle &=\left\langle2x-1,1-2y\right\rangle \\ \left.\vec\nabla f\right|_P &=\left\langle5,-15\right\rangle \\ \end{aligned}\] The directional derivative uses the unit vector. Since \(|\vec v|=\dfrac{\sqrt{5}}{3}\), the unit vector is: \[ \hat v =\dfrac{\vec v}{|\vec v|} =\left\langle \dfrac{2}{\sqrt{5}},\dfrac{1}{\sqrt{5}}\right\rangle \] So the directional derivative is: \[\begin{aligned} \left.\nabla_{\hat v}f\right|_P &=\hat v\cdot\left.\vec\nabla f\right|_P =\left\langle\dfrac{2}{\sqrt{5}},\dfrac{1}{\sqrt{5}}\right\rangle \cdot\langle5,-15\rangle \\ &=\dfrac{10}{\sqrt{5}}-\dfrac{15}{\sqrt{5}}=-\,\dfrac{5}{\sqrt{5}} =-\sqrt{5} \end{aligned}\]

The altitude of a mountain range is given by \[ a=x\sin y+y\cos x \] Find the slope of the land at \((x,y)=\left(\dfrac{\pi}{3},\dfrac{\pi}{6}\right)\) in the direction of the vector \(\vec v=\langle 4,-3\rangle\).

\(\text{slope}=\dfrac{1}{10}-\dfrac{\pi\sqrt{3}}{6}\)

We want the slope of the altitude function \(f(x,y)=x\sin y+y\cos x\) at the point \(P=\left(\dfrac{\pi}{3},\dfrac{\pi}{6}\right)\) in the direction of the vector \(\vec v=\langle 4,-3\rangle\). We calculate the gradient of the altitude and evaluate it at the point: \[\begin{aligned} &\left.\vec\nabla a\right. =\left\langle \sin y-y\sin x,x\cos y+\cos x \right\rangle \\ &\left.\vec\nabla a\right|_{(\pi/3,\pi/6)} =\left\langle\sin\dfrac{\pi}{6}-\dfrac{\pi}{6}\sin\dfrac{\pi}{3}, \dfrac{\pi}{3}\cos \dfrac{\pi}{6}+\cos \dfrac{\pi}{3} \right\rangle \\ &\qquad\qquad\quad=\left\langle \dfrac{1}{2}-\dfrac{\pi}{6}\dfrac{\sqrt{3}}{2}, \dfrac{\pi}{3}\dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\right\rangle \\ &\qquad\qquad\quad=\left\langle \dfrac{1}{2}-\dfrac{\pi\sqrt{3}}{12}, \dfrac{1}{2}+\dfrac{\pi\sqrt{3}}{6}\right\rangle \end{aligned}\] Because the slope is the directional derivative, we need to take the derivative along the unit vector. The length of \(\vec v\) is \(|\vec v|=\sqrt{4^2+3^2}=5\). So the unit vector is \(\hat v=\left\langle \dfrac{4}{5},-\,\dfrac{3}{5}\right\rangle\). So the slope is: \[\begin{aligned} \text{slope}&=\hat v\cdot\left.\vec\nabla a\right|_{(\pi/3,\pi/6)} \\ &=\left\langle \dfrac{4}{5},-\,\dfrac{3}{5}\right\rangle \cdot\left\langle \dfrac{1}{2}-\dfrac{\pi\sqrt{3}}{12}, \dfrac{1}{2}+\dfrac{\pi\sqrt{3}}{6}\right\rangle \\[8pt] &=\dfrac{4}{5}\left(\dfrac{1}{2}-\dfrac{\pi\sqrt{3}}{12}\right) -\dfrac{3}{5}\left(\dfrac{1}{2}+\dfrac{\pi\sqrt{3}}{6}\right) \\ &=\dfrac{1}{10}-\dfrac{\pi\sqrt{3}}{6} \end{aligned}\]

nx
Ham Duet is aboard the Centurian Eagle at the point \(P=(10,-5,15)\) in the middle of a deadly polaron field whose density is \(D=(x-12)^2+(y+7)^2+(z-13)^2\). Find the derivative of the density in the direction of the vector \(\vec u=\langle-2,1,2\rangle\).

\(\nabla_{\hat u}D|_P=\dfrac{20}{3}\)

The directional derivative is \(\nabla_{\hat u}D|_P=\hat u\cdot\vec\nabla D|_P\). First, we find the gradient of \(D\): \[ \vec\nabla D=\left\langle2(x-12),2(y+7),2(z-13)\right\rangle \] At \(P=(10,-5,15)\), this is \[ \vec\nabla D|_P=\langle-4,4,4\rangle \] Next, we need to find the unit vector. The length of \(\vec u=\langle-2,1,2\rangle\) is: \[ |\vec u|=\sqrt{4+1+4}=3 \] So the unit vector is \[ \hat u=\dfrac{\vec u}{|\vec u|} =\left\langle \dfrac{-2}{3},\dfrac{1}{3},\dfrac{2}{3}\right\rangle \] So the directional derivative is: \[\begin{aligned} \nabla_{\hat u}D|_P&=\hat u\cdot\vec\nabla D|_P \\ &=\left\langle \dfrac{-2}{3},\dfrac{1}{3},\dfrac{2}{3}\right\rangle \cdot\langle-4,4,4\rangle &=\dfrac{8}{3}+\dfrac{4}{3}+\dfrac{8}{3}=\dfrac{20}{3} \end{aligned}\]

d. Properties of the Gradient

The temperature in \(^\circ K\) in a room is given by \(T=\dfrac{\sin(x+y)}{\cos(x-y)}\).
1. Find the unit vector direction in which the temperature increases most rapidly at \(P=(3,8)\).
  
  \(\hat w=\langle 0.706,0.708\rangle\)
  
  We first must calculate the gradient of \(T\). To do this, we first rewrite the tempreature as: \[ T=\sec(x-y)\sin(x+y) \] Then the partial derivatives are: \[\begin{aligned} \dfrac{\partial T}{\partial x} &=\sec(x-y)\cos(x+y)+\sin(x+y)\sec(x-y)\tan(x-y) \\ \dfrac{\partial T}{\partial y} &=\sec(x-y)\cos(x+y)-\sin(x+y)\sec(x-y)\tan(x-y)) \end{aligned}\] Then the gradient at \(P=(3,8)\) is: \[\begin{aligned} \vec w&=\left.\vec\nabla T\right|_P =\left.\left\langle \dfrac{\partial T}{\partial x}, \dfrac{\partial T}{\partial y}\right\rangle\right|_P \\ &=\langle \sec(3-8)\cos(3+8)+\sin(3+8)\sec(3-8)\tan(3-8), \\ &\qquad\sec(3-8)\cos(3+8)-\sin(3+8)\sec(3-8)\tan(3-8)\rangle \\ &=\langle \sec(5)\cos(11)-\sin(11)\sec(5)\tan(5), \\ &\qquad\sec(5)\cos(11)+\sin(11)\sec(5)\tan(5)\rangle \\ &=\langle -11.902,11.933\rangle \end{aligned}\] By Property \(1\), the gradient points in the direction of \(\vec w\). Therefore, we need the unit vector of \(\vec w\). Since the length of \(\vec w\) is: \[ |\vec w|=\sqrt{11.902^2+11.933^2}=16.854 \] the unit vector in which the temperature increases most rapidly is: \[ \hat w =\left\langle \dfrac{-11.902}{16.854},\dfrac{11.933}{16.854}\right\rangle =\langle 0.706,0.708\rangle \]
2. Find the maximum rate of increase of the temperature at \(P=(3,8)\)
  
  \(\left.\nabla_{\hat w} T\right|_P =16.854\)
  
  By Property \(2\), the maximum rate of increase of the temperature is the length of the gradient: \[ \left.\nabla_{\hat w} T\right|_P =|\vec w|=16.854 \]
Consider the function \(f=\dfrac{y}{x^2}\) considered in a previous problem.
1. Find the unit vector direction in which\(f\) decreases most rapidly at \(P=(2,3)\).
  
  \(\hat w =\left\langle \dfrac{3}{\sqrt{10}},-\,\dfrac{1}{\sqrt{10}}\right\rangle\)
  
  As shown in the previous problem, the gradient of \(f\) is \(\vec\nabla f =\left\langle -\,\dfrac{3}{4},\dfrac{1}{4}\right\rangle\). By Property \(1\) the direction of maximum decrease is the negative of the gradient, \(\vec w=-\vec\nabla f =\left\langle \dfrac{3}{4},-\,\dfrac{1}{4}\right\rangle\). We need the unit vector of \(\vec w\). Since \[ |\vec w|=\sqrt{\left(\dfrac{3}{4}\right)^2+\left(\dfrac{1}{4}\right)^2} =\dfrac{\sqrt{10}}{4} \] the unit vector in which \(f\)decreases most rapidly is: \[ \hat w =\dfrac{\vec w}{|\vec w|} =\dfrac{4}{\sqrt{10}}\left\langle \dfrac{3}{4},-\,\dfrac{1}{4}\right\rangle =\left\langle \dfrac{3}{\sqrt{10}},-\,\dfrac{1}{\sqrt{10}}\right\rangle \]
2. Find the maximum rate of decrease of \(f\) at \(P=(2,3)\).
  
  \(\left.\nabla_{\hat w} T\right|_P =-\dfrac{\sqrt{10}}{4}\)
  
  By Property \(2\), the maximum rate of decrease of \(f\) is the negative of the length of the gradient: \[ \left.\nabla_{\hat w} T\right|_P =-|\vec w|=-\dfrac{\sqrt{10}}{4} \]
An bird is flying through the region where the temperaturere is \(T=275+\sin(\pi x)\cos(\pi y)+\sin(\pi z)\) in \(^\circ K\) and is currently at the point \(P=(1,1,1)\).
1. Find the unit vector direction in which the bird feels the temperature increase the fastest at \(P\).
  
  \(\hat w=\left\langle \dfrac{1}{\sqrt{2}},0,\dfrac{-1}{\sqrt{2}}\right\rangle\)
  
  We must calculate the gradient of \(T\) by finding the partial derivatives: \[\begin{aligned} \dfrac{\partial T}{\partial x}&=\pi\cos(\pi x)\cos(\pi y) \\ \dfrac{\partial T}{\partial y}&=-\pi\sin(\pi x)\sin(\pi y) \\ \dfrac{\partial T}{\partial z}&=\pi\cos(\pi z) \end{aligned}\] Then the gradient at \(P=(1,1,1)\) is: \[\begin{aligned} \vec w &=\left.\vec\nabla T\right |_P =\left .\left\langle \dfrac{\partial T}{\partial x}, \dfrac{\partial T}{\partial y}, \dfrac{\partial T}{\partial z}\right\rangle\right |_P \\ &=\langle\pi\cos\left(\pi\right)\cos(\pi), -\pi\sin\left(\pi\right)\sin(\pi), \pi\cos\left(\pi\right)\rangle \\ &=\langle \pi,0,-\pi\rangle \end{aligned}\] By property \(1\), the gradient points in the direction of \(\vec w\). Therefore, we need the unit vector of \(\vec w\). Since the length of \(\vec w\) is: \[ |\vec w|=\sqrt{\pi^2+0+\pi^2}=\pi\sqrt{2} \] The unit vector in which the temperature increases most rapidly is: \[\begin{aligned} \hat w=\left\langle \dfrac{\pi}{\pi\sqrt{2}},\dfrac{0}{\pi\sqrt{2}},\dfrac{-\pi}{\pi\sqrt{2}}\right\rangle &=\left\langle \dfrac{1}{\sqrt{2}},0,\dfrac{-1}{\sqrt{2}}\right\rangle \end{aligned}\]
  
  nx
2. Find the maximum rate of increase of the temperature at \(P\).
  
  \(\left.\nabla_{\hat w} T\right|_P=\pi\sqrt{2}\)
  
  By Property \(2\), the maximum rate of increase of the temperature is the length of the gradient: \[ \left.\nabla_{\hat w} T\right|_P =|\vec w|=\pi\sqrt{2} \]
  
  nx
  
  nx
A spacecraft is located at \((4,3,2)\) and is experiencing strong aether winds whose density is given by \(D=x^2+y^2-z^3\).
1. Find the unit vector direction in which the aether decreases most rapidly at \((4,3,2)\).
  
  \( \hat w=\left\langle\dfrac{-4}{\sqrt{61}}, \dfrac{-3}{\sqrt{61}}, \dfrac{6}{\sqrt{61}}\right\rangle \)
  
  We first calculate the gradient: \[\begin{aligned} \vec\nabla f &=\left\langle\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z}\right\rangle \\ &=\left\langle 2x,2y,-3z^2\right\rangle \\ \end{aligned}\] The gradient at \(P=(4,3,2)\) is: \[ \left.\vec\nabla f\right|_P =\left\langle 8,6,-12\right\rangle \] By Property \(1\), the direction of maximum decrease is opposite to the gradient: \[ \vec w=-\left.\vec\nabla f\right|_P =\left\langle -8,-6,12\right\rangle \] Therefore, we need the unit vector of \(\vec w\). Since the length of \(\vec w\) is: \[ |\vec w|=\sqrt{8^2+6^2+12^2}=\sqrt{64+36+144} =\sqrt{244}=2\sqrt{61} \] the direction in which the unit vector decreases most rapidly is: \[ \hat w =\dfrac{\vec w}{|\vec w|} =\dfrac{1}{2\sqrt{61}}\left\langle -8,-6,12\right\rangle =\left\langle\dfrac{-4}{\sqrt{61}}, \dfrac{-3}{\sqrt{61}}, \dfrac{6}{\sqrt{61}}\right\rangle \]
2. Find the maximum rate of decrease of the aether winds at \((4,3,2)\).
  
  \(|\vec w|=-2\sqrt{61}\)
  
  By Property \(2\), the maximum rate of decrease of the aether is the negative length of the gradient: \[ \left.\nabla_{\hat w} T\right|_P =-|\vec w|=-2\sqrt{61} \]
Find a parametric equation for the line perpendicular to the curve \(xy^2+yx^2=6\) at the point \(P=(1,2)\).

\((x,y)=(1+8t,2+5t)\)

We first calculate the gradient of \(f=xy^2+yx^2\) and evaluate it at the point \(P=(1,2)\): \[\begin{aligned} \vec\nabla f &=\left\langle\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}\right\rangle \\ &=\left\langle y^2+2yx,2xy+x^2\right\rangle \\ \left.\vec\nabla f\right|_P &=\left\langle 8,5\right\rangle \end{aligned}\] Since we want the perpendicular line, its direction is the gradient at the point, \(\vec v=\left\langle 8,5\right\rangle\). Then the parametric equation for the line is \(X=P+t\vec v\). We plug in: \[ (x,y)=(1,2)+t\left\langle 8,5\right\rangle =(1+8t,2+5t) \]
Find a non-parametric equation for the line tangent to the curve \(x^2y^2-xy=30\) at the point \(P=(2,3)\).

\(3x+2y=12\)

We first calculate the gradient of \(f=x^2y^2-xy\) and evaluate it at the point \(P=(2,3)\): \[\begin{aligned} \vec\nabla f &=\left\langle\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right\rangle \\ &=\left\langle2xy^2-y,2x^2y-x\right\rangle \\ \left.\vec\nabla f\right|_P &=\left\langle 36-3,24-2\right\rangle \\ &=\left\langle 33,22\right\rangle \end{aligned}\] The non-parametric equation of a line in \(\mathbb{R}^2\) is \(\vec n\cdot X=\vec n\cdot P\) where \(\vec n\) is the normal \(\vec n=\left.\vec\nabla f\right|_P=\left\langle 33,22\right\rangle\). So the equation is: \[ 33x+22y=33\cdot2+22\cdot3=132 \] or \[ 3x+2y=12 \]
Find the non-parametric equation for the plane tangent to the surface \(x^2z+z^2y-xy=19\) at the point \(P=(1,2,3)\).

\(4x+8y+13z=59\)

We first calculate the gradient of \(f=x^2z+z^2y-xy\) and evaluate it at the point \(P=(1,2,3)\): \[\begin{aligned} \vec\nabla f&=\left\langle \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y} ,\dfrac{\partial f}{\partial z}\right\rangle \\ &=\langle 2xz-y,z^2-x,x^2+2yz\rangle \\ \left .\vec\nabla f\right |_P&=\langle 2\cdot 3-2, 3^2-1,1+2\cdot2\cdot 3\rangle \\ &=\langle 4,8,13\rangle \end{aligned}\] The non-parametric equation of a line in \(\mathbb{R}^3\) is \(\vec N\cdot X=\vec N\cdot P\) where \(P=(1,2,3)\) is a point and \(\vec N=\left.\vec\nabla f\right|_P=\langle 4,8,13\rangle\) is the normal. Therefore, the equation is: \[ 4x+8y+13z=4+16+39=59 \]
Find a parametric equation for the line perpendicular to the surface \(xy+yz^2+xz=13\) at the point \(P=(3,1,2)\).

\((x,y,z)=(3+3t,1+7t,2+7t)\)

We first calculate the gradient of \(f=xy+yz^2+xz\) and evaluate it at point \(P=(3,1,2)\): \[\begin{aligned} \vec\nabla f&=\left\langle \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z}\right\rangle \\ &=\langle y+z,x+z^2,2yz+x\rangle \\ \left.\vec\nabla f\right|_P&=\langle 3,7,7\rangle \end{aligned}\] Since we want the perpendicular line, its direction is the gradient at the point, \(\vec v=\left.\vec\nabla f\right|_P=\langle 3,7,7\rangle\). Then the parametric equation for the line is \(X=P+t\vec v\). We plug in: \[ (x,y,z)=(3,1,2)+t\langle 3,7,7\rangle=(3+3t,1+7t,2+7t) \]
Find all points \(P=(p,q)\) on the parabola \(y=-\dfrac{1}{2}x^2+3\) where the normal vector is parallel to the vector \(\vec u=\langle4,2\rangle\).

\(P=(2,1)\).

We first rearrange the equation into the form \(y+\dfrac{1}{2}x^2=3\). We then calculate the gradient of \(f=y+\dfrac{1}{2}x^2\) and evaluate it at \(P=(p,q)\): \[\begin{aligned} \vec\nabla f &=\left\langle \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}\right\rangle \\ &=\langle x,1\rangle \\ \left .\vec\nabla f\right|_P &=\langle p, 1\rangle \end{aligned}\] We want this to be parallel to \(\vec u=\langle4,2\rangle\). So we solve the equation: \[\begin{aligned} \left .\vec\nabla f\right|_P&=\lambda\vec u \\ \langle p, 1\rangle&=\lambda\langle4,2\rangle \end{aligned}\] or \[\begin{aligned} p=4\lambda &\qquad 1=2\lambda \\ \lambda=\dfrac{1}{2} &\qquad p=2 \end{aligned}\] Then, because \(\dfrac{1}{2}p^2+q=3\), we conclude: \[ q=3-\dfrac{1}{2}p^2=3-2=1 \] Therefore, the point is \(P=(2,1)\).

We check by plugging \((p,q)=(2,1)\) into the parabola and into the normal vector to see the point is on the parabola and is parallel to \(u\). We get \[ y=-\dfrac{1}{2}x^2+3=-\dfrac{1}{2}2^2+3=1 \] and \[ \left .\vec\nabla f\right|_{(p,q)}=(2,1)=\dfrac{1}{2}\vec u \] So we know this is the correct solution.
Find all points \(P=(p,q)\) on the parabola \(y=4-x^2\) where the normal line passes through the origin \(O=(0,0)\).

\( (p,q)=(0,4) \quad \text{and} \quad (p,q)=\left(\pm\sqrt{\dfrac{7}{2}},\dfrac{1}{2}\right) \)

We first rewrite the equation as \(y+x^2=4\). We calculate the gradient of \(f=y+x^2\): \[ \vec\nabla f =\left\langle\dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y}\right\rangle =\langle 2x,1\rangle \] The direction of the normal line is the gradient evaluated at the point \(P=(p,q)\): \[ \vec v=\left.\vec\nabla f\right|_P=\langle 2p,1\rangle \] So the normal line is: \[\begin{aligned} X&=P+t\vec v=(p,q)+t\langle 2p,1\rangle \\ (x,y)&=(p+2pt,q+t) \end{aligned}\] We want this to pass through the origin \((0,0)\). So we solve: \[ (0,0)=(p+2pt,q+t) \] or \[\begin{aligned} 0=p+2pt \quad &\text{and} \quad 0=q+t \\ t=-q \quad &\text{so} \quad 0=p-2pq=p(1-2q) \end{aligned}\] So either \(p=0\) or \(q=\dfrac{1}{2}\). We also know \((p,q)\) lies on the parabola. So \(q=4-p^2\). So if \(p=0\) then \(q=4\). And if \(q=\dfrac{1}{2}\) then \(p^2=4-q=4-\dfrac{1}{2}=\dfrac{7}{2}\) or \(p=\pm\sqrt{\dfrac{7}{2}}\). Therefore the three solutions are \[ (p,q)=(0,4) \quad \text{and} \quad (p,q)=\left(\pm\sqrt{\dfrac{7}{2}},\dfrac{1}{2}\right) \]

We check the \(3\) points are on the parabola and and the normal line passes throught the origin.

\((p,q)=(0,4)\): Check the point is on the parabola: \[ y+x^2=4-0^2=4 \] The normal line is: \[ (x,y)=(p+2pt,q+t)=(0,4+t) \] which passes through \((0,0)\) when \(t=-4\).

\((p,q)=\left(\pm\sqrt{\dfrac{7}{2}},\dfrac{1}{2}\right)\): Check the point is on the parabola: \[ y+x^2=\dfrac{1}{2}+\dfrac{7}{2}=4 \] The normal line is: \[ (x,y)=(p+2pt,q+t)=\left(\pm\sqrt{\dfrac{7}{2}}(1+2t),\dfrac{1}{2}+t\right) \] which passes through \((0,0)\) when \(t=-\dfrac{1}{2}\).
Find all points \(P=(p,q,r)\) on the ellipsoid \(36x^2+9y^2+4z^2=108\) where the normal vector is parallel to the vector \(\vec v=\langle12,6,4\rangle\).

\((p,q,r)=(1,2,3) \quad \text{and} \quad (p,q,r)=(-1,-2,-3)\)

We first calculate the gradient of \(f=36x^2+9y^2+4z^2\) and evaluate it at \(P=(p,q,r)\): \[\begin{aligned} \vec\nabla f&=\left\langle \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z}\right\rangle \\ &=\langle 72x,18y,8z\rangle \\ \left.\vec\nabla f\right |_P &=\langle72p,18q,8r\rangle \end{aligned}\] We want this to be parallel to \(\vec v=\langle12,6,4\rangle\). So we need to solve the equations: \[ \langle72p,18q,8r\rangle =\lambda\langle12,6,4\rangle \] together with the equation that \(P\) lies on the ellipsoid. So the four equations are: \[\begin{aligned} 72p=\lambda12 \qquad 18&q=\lambda6 \qquad 8r=\lambda4 \\ 36p^2+9q^2&+4r^2=108 \end{aligned}\] So the first \(3\) equations say: \[ p=\lambda\dfrac{1}{6} \qquad q=\lambda\dfrac{1}{3} \qquad r=\lambda\dfrac{1}{2} \] So the ellipse equation says: \[ \lambda^2\left(36\dfrac{1}{36}+9\dfrac{1}{9}+4\dfrac{1}{4}\right)=108 \] Therefore \(\lambda^2=36\) and \(\lambda=\pm6\). This gives the two points: \[ (p,q,r)=(1,2,3) \quad \text{and} \quad (p,q,r)=(-1,-2,-3) \]

We plug the points \((p,q,r)=(1,2,3)\) and \((p,q,r)=(-1,-2,-3)\) into the ellipse equation, \(36x^2+9y^2+4z^2=108\), and into the normal vector, \(\left.\vec\nabla f\right |_P=\langle72\cdot p,18\cdot q,8\cdot r\rangle\) to see the points lie on the ellipse and the normal is parallel to \(\langle12,6,4\rangle\). For the point \((x,y,z)=\pm (1,2,3)\) we get \[ 36\cdot1^2+9\cdot2^2+4\cdot3^2=36+36+36=108 \] The normal vector is \[ \left.\vec\nabla f\right |_P=\pm \langle72\cdot1,18\cdot2,8\cdot3\rangle =\pm \langle72,36,24\rangle=\pm 6\langle12,6,4\rangle \] So we know that we have the correct answers.
Find all points \(P=(p,q,r)\) on the paraboloid \(z=5-\dfrac{x^2}{4}-\dfrac{y^2}{6}\) where the normal line passes through the origin \(O=(0,0,0)\).

There are \(5\) points: \[\begin{aligned} (p,q,r)&=(0,0,5) \\ (p,q,r)&=(0,\pm2\sqrt{3},3) \\ (p,q,r)&=(\pm2\sqrt{3},0,2) \end{aligned}\]

We first calculate the gradient of \(f=z+\dfrac{x^2}{4}+\dfrac{y^2}{6}\) \[ \vec\nabla f =\left\langle \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z}\right\rangle =\left\langle\dfrac{x}{2},\dfrac{y}{3},1\right\rangle \] The direction of the normal line is the gradient evaluated at the point \(P=(p,q,r)\): \[ \vec v=\left.\vec\nabla f\right |_P =\left\langle\dfrac{p}{2},\dfrac{q}{3},1\right\rangle \] Therefore, the normal line is: \[\begin{aligned} X&=P+t\vec v=(p,q,r)+t\left\langle\dfrac{p}{2},\dfrac{q}{3},1\right\rangle \\ &=\left(p\left(1+\dfrac{t}{2}\right),q\left(1+\dfrac{t}{3}\right),r+t\right) \end{aligned}\] We want this to pass through the point \(O=(0,0,0)\). So we need to solve the equations: \[ \left(p\left(1+\dfrac{t}{2}\right),q\left(1+\dfrac{t}{3}\right),r+t\right)=(0,0,0) \] together with the equation that \(P\) lies on the ellipsoid. So the four equations are: \[\begin{aligned} p\left(1+\dfrac{t}{2}\right)=0 \qquad q\left(1+\dfrac{t}{3}\right)&=0 \qquad r+t=0 \\ r+\dfrac{p^2}{4}+\dfrac{q^2}{6}=5 \end{aligned}\] The third equation say \(r=-t\). The first two equations lead to three cases:

a) \(p=0\) and \(q=0\): Then the ellipsoid equation says \(r=5\). So the point is \((p,q,r)=(0,0,5)\) which is the vertex of the ellipsoid. Not surprising!

b) \(p=0\) and \(t=-3\): Then \(r=3\) and the ellipsoid equation \(r+\dfrac{p^2}{4}+\dfrac{q^2}{6}=5\) says \[ 3+\dfrac{q^2}{6}=5 \quad \text{or} \qquad q=\pm\sqrt{12}=\pm2\sqrt{3} \] So there are two points \((p,q,r)=(0,\pm2\sqrt{3},3)\).

c) \(q=0\) and \(t=-2\): Then \(r=2\) and the ellipsoid equation \(r+\dfrac{p^2}{2a}+\dfrac{q^2}{2b}=5\) says \[ 2+\dfrac{p^2}{4}=5 \quad \text{or} \qquad p=\pm\sqrt{12}=\pm2\sqrt{3} \] So there are two points \((p,q,r)=(\pm2\sqrt{3},0,2)\)

We check the \(5\) points are on the paraboloid and and the normal line passes throught the origin.

\((p,q,r)=(0,0,5)\): Check the point is on the paraboloid: \[ r+\dfrac{p^2}{4}+\dfrac{q^2}{6}=5+0+0=5 \] The normal line is: \[\begin{aligned} X&=\left(p\left(1+\dfrac{t}{2}\right),q\left(1+\dfrac{t}{3}\right),r+t\right) \\ &=(0,0,r+t) \\ \end{aligned}\] which pass through \((0,0,0)\) when \(t=-5\).

\((p,q,r)=(0,\pm2\sqrt{3},3)\): Check the point is on the paraboloid: \[ r+\dfrac{p^2}{4}+\dfrac{q^2}{6}=3+0+\dfrac{12}{6}=5 \] The normal line is: \[\begin{aligned} X&=\left(p\left(1+\dfrac{t}{2}\right),q\left(1+\dfrac{t}{3}\right),r+t\right) \\ &=(0,\pm2\sqrt{3}\left(1+\dfrac{t}{3}\right),3+t) \\ \end{aligned}\] which pass through \((0,0,0)\) when \(t=-3\).

\((p,q,r)=(\pm2\sqrt{3},0,2)\): Check the point is on the paraboloid: \[ r+\dfrac{p^2}{4}+\dfrac{q^2}{6}=2+\dfrac{12}{4}+0=5 \] The normal line is: \[\begin{aligned} X&=\left(p\left(1+\dfrac{t}{2}\right),q\left(1+\dfrac{t}{3}\right),r+t\right) \\ &=(\pm2\sqrt{3}\left(1+\dfrac{t}{2}\right),0,2+t) \\ \end{aligned}\] which pass through \((0,0,0)\) when \(t=-2\).

PY: Fix alignment of list in solution.

e. Tangent and Normal, Lines and Planes to Level Sets

In \(\mathbb R^3\), consider the point \(P=(3,2,1)\) on the level surface \(F(x,y,z)=x+\dfrac{y^2}{2}+\dfrac{z^3}{3}=\dfrac{16}{3}\). Calculate each of the following:
1. Normal vector at the point \(P\).
  
  \(\vec N=\left .\vec\nabla F\right |_{P}=\langle 1,2,1\rangle\)
  
  We calculate the gradient of \(F\) and evaluate it at the point \(P\) to get the normal vector: \[\begin{aligned} \vec\nabla F&=\left\langle \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z}\right\rangle \\ &=\langle 1,y,z^2\rangle \\ \vec N=\left .\vec\nabla F\right |_{P}&=\langle 1,2,1\rangle \end{aligned}\]
2. Tangent plane at the point \(P\).
  
  \(x+2y+z=8\)
  
  The equation of a plane is \(\vec N\cdot X=\vec N\cdot P\). Since \(\vec N=\langle 1,2,1\rangle\) and \(P=(3,2,1)\), the plane is: \[ x+2y+z=8 \]
3. Normal line at the point \(P\).
  
  \((x,y,z)=(3+t,2+2t,1+t)\)
  
  The equation of a line is \(X=P+t\vec v\). For the normal line, the direction vector is the normal. Here \(\vec v=\vec N=\langle 1,2,1\rangle\) and \(P=(3,2,1)\), So the normal line is: \[\begin{aligned} X&=(3,2,1)+t\langle 1,2,1\rangle \\ (x,y,z)&=(3+t,2+2t,1+t) \\ \end{aligned}\]
In \(\mathbb R^3\), consider the point \(P=(2,2,2)\) on the level surface \(F(x,y,z)=(xy)^2\dfrac{z}{8}=4\).
1. Normal vector at point \(P\).
  
  \(\vec N=\langle 4,4,2\rangle\)
  
  We calculate the gradient of \(F\) and evaluate it at the point \(P\) to get the normal vector: \[\begin{aligned} \vec\nabla F&=\left\langle \dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}, \dfrac{\partial f}{\partial z}\right\rangle \\ &=\langle \dfrac{xy^2z}{4},\dfrac{x^2yz}{4},\dfrac{(xy)^2}{8}\rangle \\ \vec N=\left .\vec\nabla F\right |_{P}&=\langle 4,4,2\rangle \end{aligned}\]
2. Tangent plane at point \(P\).
  
  \(4x+4y+2z=20\)
  
  The equation of a plane is \(\vec N\cdot X=\vec N\cdot P\). Since \(\vec N=\langle 4,4,2\rangle\) and \(P=(2,2,2)\), the plane is: \[ 4x+4y+2z=20 \]
3. Normal line at point \(P\).
  
  \((x,y,z)=(2+4t,2+4t,2+2t)\)
  
  The equation of a line is \(X=P+t\vec v\). For the normal line, the direction vector is the normal. Here \(\vec v=\vec N=\langle 4,4,2\rangle\) and \(P=(2,2,2)\), So the normal line is: \[\begin{aligned} X&=(2,2,2)+t\langle 4,4,2\rangle \\ (x,y,z)&=(2+4t,2+4t,2+2t) \\ \end{aligned}\]
In \(\mathbb R^2\), consider the point \(P=(3,1)\) on the level curve \(F(x,y)=x^2+xy+y^3=13\).
1. Normal vector at point \(P\).
  
  \(\vec n=\langle 7,6\rangle\)
  
  We calculate the gradient of \(F\) and evaluate it at the point \(P\) to get the normal vector: \[\begin{aligned} \vec\nabla F&=\left\langle\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right\rangle \\ &=\langle 2x+y,x+3y^2,\rangle \\ \vec n=\left .\vec\nabla F\right |_{P}&=\langle 7,6\rangle \end{aligned}\]
2. Normal line at point \(P\).
  
  \(7x+6y=27\)
  
  The non-parametric equation of a line is \(\vec n\cdot X=\vec n\cdot P\). Since \(\vec n=\langle 7,6\rangle\) and \(P=(3,1)\), the plane is: \[ 7x+6y=27 \]
3. Tangent line at point \(P\).
  
  \((x,y)=(3+7t,1+6t)\)
  
  The parametric equation of a line is \(X=P+t\vec v\). For the normal line, the direction vector is the normal. Here \(\vec v=\vec n=\langle 7,6\rangle\) and \(P=(3,1)\). So the normal line is: \[\begin{aligned} X&=(3,1)+t\langle 7,6\rangle \\ (x,y)&=(3+7t,1+6t) \\ \end{aligned}\]
In \(\mathbb R^2\), consider the point \(P=(1,2)\) on the level curve \(F(x,y)=x^6+y^4+1=18\).
1. Normal vector at point \(P\).
  
  \(\vec n=\langle 6,32\rangle\)
  
  We calculate the gradient of \(F\) and evaluate it at the point \(P\) to get the normal vector: \[\begin{aligned} \vec\nabla F&=\left\langle\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right\rangle \\ &=\langle 6x^5,4y^3,\rangle \\ \vec n=\left .\vec\nabla F\right |_{P}&=\langle 6,32\rangle \end{aligned}\]
2. Normal line at point \(P\). Give a parametric and non-parametric form.
  
  \((x,y)=(1+6t,2+32t)\)
  \(y=\dfrac{16}{3}x-\dfrac{10}{3}\)
  
  The parametric equation of a line is \(X=P+t\vec v\). For the normal line, the direction vector is the normal. Here \(\vec v=\vec n=\langle 6,32\rangle\) and \(P=(1,2)\). So the normal line is: \[\begin{aligned} X&=(1,2)+t\langle 6,32\rangle \\ (x,y)&=(1+6t,2+32t) \\ \end{aligned}\] To get the non-parametric equation, we solve the equations \(x=1+6t\) and \(y=2+32t\) for \(t\) and equate them: \[ t=\dfrac{x-1}{6}=\dfrac{y-2}{32} \] Solving for \(y\), we get: \[ y-2=\dfrac{16}{3}(x-1) \] or \[ y=\dfrac{16}{3}x-\dfrac{10}{3} \]
3. Tangent line at point \(P\).
  
  \(6x+32y=70\)
  
  The non-parametric equation of a line is \(\vec n\cdot X=\vec n\cdot P\). Since \(\vec n=\langle 6,32\rangle\) and \(P=(1,2)\), the line is: \[ 6x+32y=70 \]
Find the tangent line to the equation \(F(x,y)=y-x^4=0\) at point \((2,16)\).

\(-32x+y=-48\)

The gradient of \(F(x,y)\) is \(\vec\nabla F=\left\langle -4x^3,1\right\rangle\). So the normal at \((2,16)\) is \(\vec n=\left.\vec\nabla F\right|_{(2,16)}=\left\langle -32,1 \right\rangle\). Hence, the tangent line is \(\vec n\cdot X=\vec n\cdot P\) or \[ -32x+y=-48 \]

jp

This tangent line can also be found by rewriting the equation as \(y=f(x)=x^4\) and using the \(1\) variable formula \(y=f(2)+f'(2)(x-2)\). We have \[ f(2)=16 \qquad f'(x)=4x^3 \qquad f'(2)=32 \] So the tangent line is \[ y=16+32(x-2)=32x-48 \]

Review Exercises

Jane Bond is riding in a speedboat through a lake which is poluted by acid whose density is given by \(\delta(x,y)=x^2y^2\dfrac{\text{gm}}{\text{m}^3}\) where \(x\) and \(y\) are measured in \(\text{m}\). Currently she is at the point \(P=(2,1)\) and wants to get out of the acid as fast as possible.
1. If Jane's current velocity is \(\vec v=\left\langle \dfrac{1}{4},\dfrac{1}{2}\right\rangle \dfrac{\text{m}}{\text{sec}}\), what is the current rate at which she sees the acid density changing? Is it increasing or decreasing?
  
  The density is increasing at \(\dfrac{\partial \delta}{\partial t}=5\).
  
  The gradient of the density and its value at the current position are: \[\begin{aligned} \vec\nabla \delta&=\langle 2xy^2,2x^2y\rangle \\ \left .\vec\nabla \delta\right |_{(2,1)}&=\langle 4,8\rangle \end{aligned}\] So the rate of change of the density is: \[\begin{aligned} \dfrac{\partial \delta}{\partial t}&=\vec v\cdot\left .\vec\nabla \delta\right|_ {(2,1)}=\left\langle \dfrac{1}{4},\dfrac{1}{2}\right\rangle\cdot\langle 4,8\rangle \\ &=1+4=5 \end{aligned}\] Since this is positive, the density is increasing.
2. Find the unit vector direction, \(\hat w\), in which the acid density decreases fastest.
  
  \(\hat w =\left\langle \dfrac{-1}{\sqrt{5}},\dfrac{-2}{\sqrt{5}}\right\rangle\)
  
  The density decreases fastest in the direction opposite to the gradient \[ \vec w=-\left .\vec\nabla \delta\right |_{(2,1)}=\langle -4,-8\rangle \] Its length is: \[ |\vec w|=\sqrt{4^2+8^2}=\sqrt{80}=4\sqrt{5} \] So the unit vector of maximum decrease is: \[ \hat w=\dfrac{\vec w}{|\vec w|} =\left\langle \dfrac{-1}{\sqrt{5}},\dfrac{-2}{\sqrt{5}}\right\rangle \]
3. Find the maximum rate of decrease of the acid density.
  (It's negative because it is decreasing.)
  
  \(\dfrac{\partial \delta}{\partial t}=\nabla_{\hat w}\delta =-4\sqrt{5}\)
  
  The maximum rate of increase is the length of the gradient which was previously found to be: \[ \left .\vec\nabla \delta\right |_{(2,1)}=4\sqrt{5} \] The maximum rate of decrease is its negative: \[ \dfrac{\partial \delta}{\partial t}=\nabla_{\hat w}\delta =-4\sqrt{5} \]
  
  To check, we directly compute the derivative along \(\hat w =\left\langle \dfrac{-1}{\sqrt{5}},\dfrac{-2}{\sqrt{5}}\right\rangle\): \[\begin{aligned} \dfrac{\partial \delta}{\partial t} &=\nabla_{\hat w}\delta =\hat w\cdot\left .\vec\nabla \delta\right |_{(2,1)}\\ &=\left\langle \dfrac{-1}{\sqrt{5}},\dfrac{-2}{\sqrt{5}}\right\rangle \cdot\langle 4,8\rangle \\ &=\dfrac{-1}{\sqrt{5}}\cdot4+\dfrac{-2}{\sqrt{5}}\cdot8 =-4\sqrt{5} \end{aligned}\]
4. If the maximum speed of her speedboat is \(3\dfrac{\text{m}}{\text{sec}}\), find the velocity, \(\vec u\), the boat should have to decrease the acid density as fast as possible.
  
  The velocity is its speed times its direction.
  
  \(\vec u =\left\langle\dfrac{-3}{\sqrt{5}},\dfrac{-6}{\sqrt{5}}\right\rangle\)
  
  The maximum speed is \(|\vec u|=3\) and the direction of maximum decrease is \[ \hat u=\hat w =\left\langle\dfrac{-1}{\sqrt{5}},\dfrac{-2}{\sqrt{5}}\right\rangle \] So the velocity of maximum decrease is: \[ \vec u=|\vec u|\hat u =\left\langle\dfrac{-3}{\sqrt{5}},\dfrac{-6}{\sqrt{5}}\right\rangle \]
5. Find the rate the acid density is changing if the speedboat has velocity \(\vec u\).
  
  \(\dfrac{\partial f}{\partial t}=\nabla_{\vec u} \delta=12\sqrt{5}\)
  
  \[ \dfrac{\partial \delta}{\partial t}=\nabla_{\vec u} \delta =|\vec u|\nabla_{\hat u}\delta=3\cdot4\sqrt{5}=12\sqrt{5} \]
A spacecraft is experiencing strong aether winds whose density is given by \(D=xy^{\small z}\).
1. Find the unit vector direction in which the aether decreases most rapidly at the point \(P=(2,e,0)\).
  
  \(\hat w =\left\langle\dfrac{-1}{\sqrt{5}},0,\dfrac{-2}{\sqrt{5}}\right\rangle\)
  
  We first calculate the gradient: \[ \vec\nabla D=\langle y^z,xzy^{z-1},\ln(y)xy^z\rangle \] The gradient at \(P=(2,e,0)\) is: \[ \left.\vec\nabla D\right|_P=\langle 1,0,2\rangle \] However, we need to find the unit vector that decreases most rapidly, so we to take the negative of this to get: \[ \vec w=\langle -1,0,-2\rangle \] Next, we compute the length of \(\vec w\). \[ |\vec w|=\sqrt{1+4}=\sqrt{5} \] So the direction which the aether decreases most rapidly is: \[ \hat w=\dfrac{\vec w}{|\vec w|} =\left\langle\dfrac{-1}{\sqrt{5}},0,\dfrac{-2}{\sqrt{5}}\right\rangle \]
2. Find the maximum rate of decrease of the aether winds at the point \(P=(2,e,0)\).
  
  \(\left .\nabla_{\hat w}D\right |_P=-\sqrt{5}\)
  
  By Property \(2\), the maximum rate of decrease of the aether is the negative length of the gradient: \[ \left .\nabla_{\hat w}D\right |_P=-|\vec w|=-\sqrt{5} \]

PY: All Checked