14. Directional Derivatives and Gradients

d. Properties of the Gradient

  1. \(\vec\nabla f\) points in the direction of maximum increase of \(f\).
  2. \(|\vec\nabla f|\) is the maximum rate of increase of \(f\).

2. Applications of Gradient Properties 1 & 2

The temperature in \(^\circ K\) in a room is given by \(T=275+\sin(\pi x)\cos(\pi y)+\sin(\pi z)\).

  1. Find the unit vector direction in which the temperature increases most rapidly at \(P=(2,-1,2)\).
  2. Find the maximum rate of increase of the temperature at \(P=(2,-1,2)\).

From an example on a previous page, the gradient of the temperature at \(P\) is: \[ \vec w=\left.\vec\nabla T\right|_P =\langle -\pi,0,\pi\rangle \]

  1. By Property 1, the gradient points in the direction of maximum increase of the function. So we need the unit vector in the direction of \(\vec w\). Since \(|\vec w|=\sqrt{\pi^2+\pi^2}=\pi\sqrt{2}\), the unit vector direction in which the temperature increases most rapidly is: \[ \hat w=\left\langle -\,\dfrac{1}{\sqrt{2}},0,\dfrac{1}{\sqrt{2}}\right\rangle \]
  2. By Property 2, the maximum rate of increase of the temperature is the length of the gradient, or \(|\vec w|=\pi\sqrt{2}\).

Duke Skywater is flying the Centurion Eagle through a deadly polaron field whose density is given by \(p(x,y,z)=xyz^2\,\dfrac{\text{radions}}{\text{millilightyear}^3}\) where \(x\), \(y\) and \(z\) are measured in \(\text{millilightyears}\). Assume Duke's current position is \((x,y,z)=(4,-2,3)\).

  1. If Duke's current velocity is \(\vec v=\langle .2,.3,.4\rangle\,\dfrac{\text{millilightyears}}{\text{year}}\), what is the current rate at which he sees the polaron density changing? Is it increasing or decreasing?

    The density is decreasing at \(\dfrac{\partial p}{\partial t}=-12\).

    The gradient of the density and its value at the current position are: \[\begin{aligned} \vec\nabla p&=\langle yz^2,xz^2,2xyz\rangle \\ \left.\vec\nabla p\right|_{(4,-2,3)}&=\langle -18,36,-48\rangle \end{aligned}\] So the rate of change of the density is: \[\begin{aligned} \dfrac{dp}{dt} &=\vec v\cdot\left.\vec\nabla p\right|_{(4,-2,3)} =\langle .2,.3,.4\rangle\cdot\langle -18,36,-48\rangle \\ &=-3.6+10.8-19.2=-12 \end{aligned}\] So the density is decreasing.

  2. Find the unit vector direction, \(\hat u\), in which the polaron density decreases fastest.

    If a quantity increases fastest in the direction \(\hat w\), then it decreases fastest in the direction \(-\hat w\).

    \(\hat u =\left\langle \dfrac{3}{\sqrt{109}},-\,\dfrac{6}{\sqrt{109}}, \dfrac{8}{\sqrt{109}}\right\rangle\)

    The density increases fastest in the direction of the gradient \[ \vec w=\left.\vec\nabla p\right|_{(4,-2,3)} =\langle -18,36,-48\rangle \] Its length is: \[ |\vec w|=\sqrt{18^2+36^2+48^2}=6\sqrt{3^2+6^2+8^2}=6\sqrt{109} \] So the unit vector of maximum increase is: \[ \hat w=\left\langle -\,\dfrac{3}{\sqrt{109}},\dfrac{6}{\sqrt{109}}, -\,\dfrac{8}{\sqrt{109}}\right\rangle \] and the unit vector of maximum decrease is: \[ \hat u=-\hat w =\left\langle \dfrac{3}{\sqrt{109}},-\,\dfrac{6}{\sqrt{109}}, \dfrac{8}{\sqrt{109}}\right\rangle \]

  3. Find the rate the polaron density is changing in the direction \(\hat u\).

    \(\dfrac{\partial p}{\partial t}=-6\sqrt{109}\)

    The maximum rate of increase is the length of the gradient which was previously found to be \[ |\vec w|=\left|\left.\vec\nabla p\right|_{(4,-2,3)}\right|=6\sqrt{109} \] So the maximum rate of decrease is the negative of this \[ \dfrac{\partial p}{\partial t}=-6\sqrt{109} \]

  4. If the maximum speed of the Centurion Eagle is \(2\) millilightyears/year, find the velocity, \(\vec u\), the Centurion Eagle should have to decrease the polaron density as fast as possible.

    The velocity is its speed times its direction.

    \(\vec u =\left\langle \dfrac{6}{\sqrt{109}},-\,\dfrac{12}{\sqrt{109}}, \dfrac{16}{\sqrt{109}}\right\rangle\)

    The maximum speed is \(|\vec u|=2\) and the direction of maximum decrease is \(\hat u =\left\langle \dfrac{3}{\sqrt{109}},-\,\dfrac{6}{\sqrt{109}}, \dfrac{8}{\sqrt{109}}\right\rangle\). So the velocity of maximum decrease is: \[ \vec u=|\vec u|\hat u =\left\langle \dfrac{6}{\sqrt{109}},-\,\dfrac{12}{\sqrt{109}}, \dfrac{16}{\sqrt{109}}\right\rangle \]

  5. Find the rate the polaron density is changing if the Centurion Eagle has velocity \(\vec u\).

    \(\dfrac{\partial p}{\partial t}=-12\sqrt{109}\)

    \[ \dfrac{\partial p}{\partial t}=\nabla_{\vec u} p=|\vec u|\nabla_{\hat u} p=2(-6\sqrt{109})=-12\sqrt{109} \]

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