Exercise 13 solutions | Ludek Cizinsky

🔹 14.6.55

If $f(x, y) = xy$, find the gradient vector $\nabla f(3, 2)$ and use it to find the tangent line to the level curve $f(x, y) = 6$ at the point $(3, 2)$. Sketch the level curve, the tangent line, and the gradient vector.

Solution

We start by finding the gradient vector. Since we know that in general $\nabla f(x, y) = \langle f_x, f_y \rangle$. Therefore, we know that for our function $f$, the gradient vector is $\nabla f(x, y) = \langle y , x \rangle$ and thus $\nabla f(3, 2) = \langle 2, 3 \rangle$.

We know that the tangent line goes through the point $P = (x_0, y_0) = (3, 2)$. Let $v$ be a tangent vector at point $P$ such that $v = \langle x - x_0, y - y_0\rangle$ for all $(x, y)$ which lie on the tangent line.

Finally, we know that gradient vector is perpendicular to the tangent vector at the point $P$, therefore, we can write $\nabla f(3, 2) \cdot v = 0$. This means that $2(x - x_0) + 3(y - y_0) = 2(x - 3) + 3(y - 2) = 0$. After doing some additional algebra we arrive to the following equation: $y = -\frac{2}{3}x + 4$.

See the visualization here.

⛑ 14.6.61

Are there any points on the hyperboloid $x^2 - y^2 - z^2 = 1$ where the tangent plane is parallel to the plane $z = x + y$?

Solution

First, we should define a tangent plane for the given surface $f(x, y, z) = x^2 - y^2 - z^2 = 1$. We start by computing the gradient vector of $f(x, y, z)$:

\[$\nabla F(x, y, z) = \langle 2x, -2y, -2z \rangle$\]

Then, let $u$ be a vector that is part of tanget plane touching the surface $f(x, y, z) = 1$ at some point $P(x_0, y_0, z_0)$:

\[u = <x - x_0, y - y_0, z - z_0>\]

Then, the tangent plane can be defined as follows:

\[\begin{aligned} \nabla f(x_0, y_0, z_0) \cdot u = 0 \\ 2x_0(x - x_0) - 2y_0(y - y_0) - 2z_0(z - z_0) = 0 \\ x_0(x - x_0) - y_0(y - y_0) - z_0(z - z_0) = 0 \\ z_0z - z_0^2 = x_0x - x_0^2 - y_0y - y_0^2 \\ z_0z = x_0x - x_0^2 - y_0y - y_0^2 + z_0^2 \\ z_0z = x_0x - y_0y - x_0^2 - y_0^2 + z_0^2 \\ z = \frac{x_0}{z_0}x - \frac{y_0}{z_0}y - \frac{x_0^2 + y_0^2 - z_0^2}{z_0} \end{aligned}\]

That was a lot of algebra, but we got the formula for the tangent plane for the given surface at some point $P(x_0, y_0)$.

Now, we know that two planes are parallel if they have the same gradient vector. Let’s compute these for each plane:

tangent plane: $<\frac{x_0}{z_0}, - \frac{y_0}{z_0}>$
given plane: $<1, 1>$

In addition, given that the point $P(x_0, y_0)$ must be live on the surface $f(x, y, z) = 1$, we construct the following system of equations:

\[\begin{aligned} 1 = \frac{x_0}{z_0} \\ 1 = - \frac{y_0}{z_0} \\ x_0^2 - y_0^2 - z_0^2 = 1 \end{aligned}\]

From the first two equations, we know that:

\[\begin{aligned} x_0 = z_0 \\ y_0 = - z_0 \end{aligned}\]

We then substitute these into the third equation:

\[\begin{aligned} z_0^2 - (-z_0)^2 - z_0^2 = 1 \\ z_0^2 - z_0^2 - z_0^2 = 1 \\ - z_0^2 = 1 \\ z_0^2 = -1 \end{aligned}\]

Clearly, this is not possible, therefore there are no such points at which the tangent plane would be parallel to given plane.

🔸 Exercise from the course

Suppose $f : R^2 \rightarrow R$ satisfies $f(1, 2) = 2$, $\nabla f(1, 2) = (2, -1)$. Estimate the value of $f(2, 3)$.

Solution

We should first notice that the points $P(1, 2)$ and $P(2, 3)$ are quite close to each other. In other words, we could perhaps use some kind of approaximation of how function $f$ behaves around the point $P(1, 2)$ in order to estimate the other given point. One of the approximation techniques which we have in our toolset after this lecture is to use a tangent plane as a proxy to $f$ at the given point $P$.

We know that a tangent plane at a particular point $P$ and towards some surface $f$, has the following formula:

\[g(x, y) = f(x_0, y_0) + f_x(x_0, y_0)(x - x_0) + f_y(x_0, y_0)(y - y_0)\]

Therefore, for the given values, our tangent plane is $g(x, y) = 2 + 2(x - 1) - (y - 2)$.

So if we plug in $P(2, 3)$, we should obtain as an approximation value 3.

🔹 14.7.7

Find the local maximum and minimum values and saddle point(s) of the function $f(x, y) = 2x^2 - 8xy + y^4 - 4y^3$.

Solution

In order to find critical points, we need to find such points $(x, y)$ where $\nabla f(x , y) = \langle 0, 0 \rangle$. Therefore, we need to first find $f_x$ and $f_y$ which is $f_x = 4x - 8y$ and $f_y = -8x +4y^3 - 12y^2$. Then we simply need to solve following system of equations:

\[\begin{aligned} 4x - 8 = 0 \\ -8x +4y^3 - 12y^2 = 0 \end{aligned}\]

If done correctly, you should obtain following critical points: $P_1(0, 0), P_2(8, 4), P_3(-2, -1)$.

Now, we need to use second derivative test to determine what kind of points we obtained. Thus, we first compute the following terms needed for the second derivative test: $f_{xx}(x, y) = 4$, $f_yy(x, y) = 12y^2 - 24y$ and $f_{xy}(x, y) = 8$.

We then use these to compute $D$:

\[D = f_{xx}f_{yy} - f_{xy}^2\]

For $P_1$, we obtain $D = -64$ which means $P_1$ is a saddle point. Similarly, for $P_2$ we obtain $D = 320$, thus it is local minimum. Finally, for $P_3$ we obtain $D = 80$ which also means it is local minimum.

See the visualization here.

🔸 14.8.5

Each of these extreme value problems has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of the function subject to the given constraint. Function is $f(x, y) = xy$ and the constraint is $4x^2 + y^2 = 8$.

Solution

We can find extreme values of $f$ under the constraint of $g(x, y) = c$ using the equation $\nabla f(x, y) = \lambda \nabla g(x, y)$. This equation says that at given extreme point (assuming it exists), the two functions will have their normal vectors parallel. $\lambda$ represents lagrange multiplier.

Therefore, we need to solve the following system of equations:

\[\begin{aligned} y = \lambda 8x \\ x = \lambda 2y \\ 4x^2 + y^2 = 8 \end{aligned}\]

This is usually the hardest part since there is no recipe on how to do this as it was for example when solving a system of linear equations. We can start by plug in $y$ from the first equation to the second equation:

\[x = \lambda^2 16x \Rightarrow x - \lambda^2 16x = 0 \Rightarrow x(1 - \lambda^2 16) = x(1 - 4\lambda)(1 + 4\lambda) = 0\]

This equation implies that if $x = 0$, then $y = 2\sqrt{3}$ (from third equation), but this would violate the first equation where if $x = 0$, then $y = 0$ as well. Thus, we can dismiss this option.

Next, if we consider the case where $\lambda = \pm \frac{1}{4}$, then according to the first and second equation, this would mean that $x = \pm 1$ and $y = \pm 2$. Both meet the third equation, thus the conclusion is that we consider combination of points with $x = \pm 1$ and $y = \pm 2$.

To find out what kind of extreme each point represents, we plug them into $f$. For points $(1, 2)$ and $(-1, -2)$, we have should have max and for the points $(-1, 2)$ and $(1, -2)$, we have a min.

⛑ 14.8.47

Find the maximum volume of a rectangular box that is inscribed in a sphere of radius $r$. To solve this problem, use Lagrange multiplier method.

Solution

First, we define our function $f$ for which we want to find maximum as $f(x, y, z) = xyz$. Next, we define the constraint $g(x, y, z) = c$. Note that $x$, $y$ and $z$ are length, width and height of the rectangle respectively. Now we need to find out how to constraint them such that the rectangle is inscribed in the sphere with radius $r$. In other words, we should think of what relationship have the parameters of $f$ with radius. It turns out that if we define diameter as $d$, then $d = x^2 + y^2 + z^2$. And indeed $d = 2r$, thus our constraint $g$ can be defined as follows $g(x, y, z) = x^2 + y^2 + z^2 = 4r^2$.

You then choose the same strategy to solve this problem as in 14.8.5. First, you should obtain that $x = y = z = \pm 2\lambda$. After this, you simply plug these values into $g(x, y, z) = 4r^2$ which yields $\lambda = \pm \frac{\sqrt{3}}{3}r$. Given the nature of $x$, $y$ and $z$ you can just write $\lambda = \frac{\sqrt{3}}{3}r$.

This implies that $x = y = z = 2\frac{\sqrt{3}}{3}r$ and volume of such rectangle would be $\frac{8r^3}{3\sqrt{3}}$.

🔍 About the page

Last updated: 01/12/2022
Unless othewise stated, exercises come from the book: James Stewart, Daniel K. Clegg and Saleem Watson: Calculus: Early Transcendentals, Metric Edition, 9th edition
Purpose: This page was created as part of a preparation for my teaching assistant session in the course Linear algebra and optimization managed by Rasmus Ejlers Møgelberg. Please note that the solutions on this page are my own, thus it should not be considered as official solutions provided as part of the course.