14. Directional Derivatives and Gradients
d. Properties of the Gradient
- \(\vec\nabla f\) points perpendicular to each level set of \(f\).
- \(|\vec\nabla f|\) is inversely proportional to the spacing between level sets. (Qualitative)
3. Applications of Gradient Properties 3 & 4
Find a normal vector to the ellipse \(\dfrac{x^2}{8}+\dfrac{y^2}{18}=1\) at the point \((x,y)=(2,3)\).
The ellipse is the level set of \(f=\dfrac{x^2}{8}+\dfrac{y^2}{18}\) with value \(1\). Its gradient will be perpendicular to the level set. The gradient is: \[ \vec\nabla f=\left\langle \dfrac{x}{4},\dfrac{y}{9}\right\rangle \] At \((x,y)=(2,3)\), the normal is: \[ \vec N=\left.\vec\nabla f\right|_{(2,3)} =\left\langle \dfrac{1}{2},\dfrac{1}{3}\right\rangle \] Any multiple of this is acceptable.
The plot at the right is a contour plot of \(f(x,y)\). At which point is \(|\vec\nabla f|\) the largest?
Point \(P\)
Point \(Q\)
Point \(R\)
Point \(S\)
![GradContourEx.jpg](graphics/GradContourEx.jpg)
\(P\). Incorrect. At this point the countours are far apart. What does that say about \(|\vec\nabla f|\)? Try again.
\(Q\). Sorry. At this point the countours are farther apart than at \(R\) but closer than at \(P\). What does that say about \(|\vec\nabla f|\)? Try again.
\(R\). Sorry. At this point the countours are farther apart than at \(S\) but closer than at \(Q\). What does that say about \(|\vec\nabla f|\)? Try again.
\(S\). Correct. At this point the countours are close together. So \(|\vec\nabla f|\) is largest.
Find the points on the hyperboloid \(2x^2-y^2+2z^2=1\) where the normal line is parallel to the line that joins the points \(P=(2,-1,3)\) and \(Q=(6,-4,1)\).
The points are: \[ (x,y,z)=(2,3,-1) \qquad \text{and} \qquad (x,y,z)=(-2,-3,1) \]
The hyperboloid is a level set of the function \(f(x,y,z)=2x^2-y^2+2z^2\). The direction of the normal line is its gradient: \[ \vec\nabla f=\langle 4x,-2y,4z\rangle \] We want this to be parallel to the vector from \(P\) to \(Q\): \[ \overrightarrow{PQ}=Q-P=\langle 4,-3,-2\rangle \] To say they are parallel means one vector is a multiple of the other, i.e. \[ \vec\nabla f=\lambda\overrightarrow{PQ} \qquad \text{or} \qquad \langle 4x,-2y,4z\rangle =\lambda\langle 4,-3,-2\rangle \] where \(\lambda\) (the Greek letter lambda) is the proportionality factor. This gives \(3\) equations, \[ 4x=4\lambda \qquad -2y=-3\lambda \qquad 4z=-2\lambda, \] but there are \(4\) unknowns: \(x\), \(y\), \(z\) and \(\lambda\). We need a \(4^\text{th}\) equation. It is the equation of the hyperboloid: \[ 2x^2-y^2+2z^2=1 \] To solve these equations, we use the first three equations to express \(x\), \(y\) and \(z\) in terms of \(\lambda\) and substitute these into the equation of the hyperbola: \[ x=\lambda \qquad y=\dfrac{3}{2}\lambda \qquad z=-\,\dfrac{1}{2}\lambda \] \[ 2(\lambda)^2-\left(\dfrac{3}{2}\lambda\right)^2 +2\left(-\,\dfrac{1}{2}\lambda\right)^2=1 \] This says \((2-\,\dfrac{9}{4}+\dfrac{1}{2})\lambda^2=1\) or \(\dfrac{\lambda^2}{4}=1\) or \(\lambda=\pm2\). So the points are: \[ (x,y,z)=(2,3,-1) \qquad \text{and} \qquad (x,y,z)=(-2,-3,1) \]
The method used to solve this problem is similar to one that will be used later to solve Max-Min problems using Lagrange multipliers.
There are more significant applications of Property 3 on the next three pages.
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