Math 8 Practice Exam Problems

As always, no guarantees on solutions...

1.

Find the general solution to the differential equation $$\frac{dy}{dx} + \frac{y}{x\ln x} = x$$.

This is a first order linear differential equation with integrating factor equal to $$e^{\int dx/(x\ln x)}$$. Now $$\int \frac{dx}{x\ln x} = \ln(\ln x) + C$$ using the substitution $u = \ln x$, so $$e^{\int dx/(x\ln x)} = e^{\ln(\ln x)} = \ln x$$. Multiplying both sides of the differential equation by this factor yields $$\frac{d}{dx}\left(y \ln x\right) = x\ln x$$. We integrate (the right hand side by parts): $$y\ln x = \int x\ln x\,dx = \frac12 x^2\ln x - \frac{x^2}{4} + C$$, hence $$y = \frac12 x^2 - \frac{x^2}{4\ln x} + \frac{C}{\ln x}$$.

2.

Find the equation of the tangent plane to the level surface of $f(x,y,z) = ye^{-x^2}\sin z$ at $(0,1,\pi/3)$. We need only compute the gradient of $f$ at $(0,1,\pi/3)$.
$$\nabla f(x,y,z) = \langle -2xye^{-x^2}\sin z,e^{-x^2}\sin z, ye^{-x^2}\cos z\rangle$$, so $$\nabla f(0,1,\pi/3) = \langle 0, \frac{\sqrt 3}{2}, \frac12 \rangle$$, and the equation of the plane is $$0(x-0) + \frac{\sqrt 3}{2}(y-1) + \frac12 (z-\pi/3) = 0$$, or more simply, $\sqrt 3 y + z = \sqrt3 + \pi/3$.

3.

Suppose that $z = f(x,y)$ is a smooth real-valued function of two variables, and that $$\frac{\partial f}{\partial x}(1,1) = 3$$ and $$\frac{\partial f}{\partial y}(1,1) = -1$$. If $x = s^2$ and $y = s^3$, we may then view $z$ as a function of the single variable $s$. The value of $$\frac{dz}{ds}$$ at $s = 1$ is:

By the chain rule, $$\frac{dz}{ds} = \frac{\partial z}{\partial x}\frac{dx}{ds} + \... ...dy}{ds} = \frac{\partial z}{\partial x}2s + \frac{\partial z}{\partial y}3s^2$$. When $s = 1$, $x = y = 1$, so we have $$\frac{dz}{ds}\Big\vert _{s=1}$$ is 3(2) + (-1)(3) = 3.

4.

Find an equation of the curve $y = f(x)$ that passes through the point $(1,1)$ and intersects all level curves of the function $g(x,y) = x^4 + y^2$ at right angles.

$\nabla g(x,y) = \langle 4x^3, 2y\rangle$ is always orthogonal to the level curves of $g(x,y)$, so we want the tangent line to $y = f(x)$ to be parallel to the gradient at each point $(x,y)$, that is, the slope $$\frac{dy}{dx} = \frac{2y}{4x^3} = \frac{y}{2x^3}$$. This is a separable differential equation: $$\frac{dy}{y} = \frac{dx}{2x^3}$$. Integrating, we obtain $\ln \vert y\vert = -x^{-2}/4 + C$, so $$y = Ae^{-x^{-2}/4}$$. Given that the curve passes through $(1,1)$, we have $$y = e^{1/4}e^{-x^{-2}/4}$$.

5.

A ball is placed at the point $(1,2,3)$ on the surface $z= y^2 - x^2$. Give the direction in the $xy$-plane corresponding to the direction in which the ball will start to roll. Describe the path in the $xy$-plane which the ball will follow. At the point $(1,2,3)$ what is the maximum rate at which the ball is descending. Let $f(x,y) = y^2 - x^2$. The gradient of $f$ points in the direction of maximum increase of $f$ (the height), so $-\nabla f$ points in the direction of maximum decrease, and hence the direction in which the ball will roll. $$\nabla f(x,y) =\langle -2x,2y\rangle$$, so $-\displaystyle \nabla f(1,2) = \langle 2,-4 \rangle$ indicates the direction in which the ball will start to roll. The path followed by the ball is similar to the previous problem, namely we expect $$\frac{dy}{dx} = \frac{-y}{x}$$. This yields $y = 2/x$ for a solution. The maximum rate of increase is $$\vert\nabla f(1,2)\vert = \sqrt{20}$$.

6.

Let $f(x,y)= x^4 + y^4 + x^2 - y^2$. Find and classify all critical points of $f$. Use the method of Lagrange multipliers to find the largest and smallest values of $f$ on the circle $x^2 + y^2 = 4$.

$$\nabla f(x,y) = \langle 4x^3 + 2x, 4y^3 - 2y\rangle = \langle 0,0 \rangle$$. From this we have $x(2x^2+1) = 0$ and $y(2y^2 - 1)$, so the only solutions are $x=0$, and $y = 0, \pm 1/\sqrt2$.

The Hessian matrix (second derivative test) is $\begin{pmatrix} 12x^2 + 2&0\\ 0&12y^2-2 \end{pmatrix}$

At the points $(0, \pm 1/\sqrt2)$, the determinant is positive indicating local extrema which, since $f_{11}(0,\pm 1/\sqrt2) > 0$, are local minima. At $(0,0)$, the Hessian has negative determinant, indicating the origin is a saddle point. For Lagrange multipliers, let $g(x,y) = x^2 + y^2$. We need to solve the system given by $\nabla f = \lambda \nabla g$ and $g(x,y) = 4$. This gives

$$4x^3 + 2x = \lambda 2x$$

$$4y^3 - 2y = \lambda 2y$$

$$x^2 + y^2 = 4$$

We have to exercise a little caution here. We would like to say that $\lambda = 2x^2 + 1 = 2y^2 - 1$, but this is true only if $x$ and $y$ are not zero. Since zero is a possible value for $x$ and $y$, we must make three cases: $x=0$ yielding the two points $(0, \pm 2)$; $y = 0$ yielding the two points $(\pm2, 0)$, and the ``desired'' case where $2x^2 + 1 = 2y^2 - 1$. In this last case, we have $x^2 - y^2 = -1$ and $x^2 + y^2 = 4$ which yields (adding the two equations) $2x^2 = 3$. $x^2 = 3/2$ implies $y^2 = 5/2$ which provides 4 points $(\pm \sqrt{3/2}, \pm \sqrt{5/2})$.

Testing all points we see that $f(0,\pm2) = 12$, $f(\pm2,0) = 20$, and $f(\pm \sqrt{3/2}, \pm \sqrt{5/2}) = 15/2$. Thus the maximum value is 20 which occurs at $(\pm2, 0)$, and the minimum value is $15/2$ which occurs at $(\pm \sqrt{3/2}, \pm \sqrt{5/2})$.

7.

Consider a function $z = f(x,y)$ which is defined and has partial derivatives of all orders for all $x$ and $y$. Suppose the function $f(x,b)$ has a local maximum at $x=a$ and the function $f(a,y)$ has a local minimum at $y=b$. Can one infer that the point $(a,b)$ is a critical point, saddle point, local maximum, local minimum?

This is clearly a saddle point as the function increases in the direction of the $y$ axis, but decreases in the direction of the $x$ axis.