Math 234 — Study guide 2nd midterm

Syllabus | Homework assignments | About the exams | Using your computer | Lecture schedule

Topics covered

The chain rule, linear approximation, and tangents to graphs and levelsets

You should understand the linear approximation formula, how it implies the chain rule, and you should know how to use the chain rule to compute derivatives of first and second order of a function.

Higher derivatives and Clairaut’s Theorem

Know that mixed partials are equal, i.e. \[ f_{xy} = f_{yx} \] (provided they are continuous). Know when two given functions $P(x,y)$ and $Q(x,y)$ can be the partial derivatives of another function $f(x,y)$, and know how to find that function $f$ when it exists.

Typical problems

Find the tangent (line or plane?) to
- graph of $z=f(x,y)$
- level set of $z=f(x,y)$
- level set of $F(x,y,z)$
In more detail, complete the following (i.e. fill in the missing words or equations):
- Given a function $z=f(x,y)$ and a point $(x_0,y_0)$ in its domain. If the point $(x_0,y_0,z_0)$ is on the graph of the function, then $z_0=$ $f(x_0,y_0)$. The equation for the tangent plane to the graph at the point $(x_0,y_0,z_0)$ is $z-z_0 = f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0) (y-y_0)$
- Given a function $z=f(x,y)$ and a point $(x_0,y_0)$ in its domain. If the point $(x_0,y_0)$ is on the levelset at level $c$ of the function, then $c=$ $f(x_0,y_0)$. The equation for the tangent line to the levelset of $f$ at the point $(x_0,y_0)$ is $\vec\nabla f(x_0,y_0) \cdot (\vx-\va) = 0$ (vector form) $f_x(x_0,y_0)(x-x_0) + f_y(x_0,y_0) (y-y_0) = 0$ (non vector form)
- Given a function $w=F(x,y,z)$ and a point $(x_0,y_0,z_0)$ in its domain. If the point $(x_0,y_0,z_0)$ is on the levelset at level $c$ of the function $F$, then $c=$ $F(x_0,y_0,z_0)$. The equation for the tangent plane to the levelset of $F$ at the point $(x_0,y_0,z_0)$ is $\vec\nabla F(x_0,y_0,z_0) \cdot (\vx-\va) = 0$ (vector form) or $F_x(x_0,y_0,z_0)(x-x_0) + F_y(x_0,y_0,z_0) (y-y_0) + F_z(x_0,y_0,z_0) (z-z_0) = 0$
Use the linear approximation to approximate $\sqrt[4]{101\cdot 9^2}$
(with the hint that $\sqrt[4]{100\cdot 10^2} = \sqrt[4]{10^4} = 10$.)
Define the function $f(x,y) = \bigl(xy^2\bigr)^{1/4} = x^{1/4}y^{1/2}$. We find the linear approximation at $x_0=100, y_0=10$ and substitute $\Delta x=1$, $\Delta y = -1$: \[ f(100+\Delta x, 10+\Delta y) = f(100, 10) + f_x(100, 10)\Delta x + f_y(100, 10)\Delta y + \text{error} \] We have $f_x = \frac14 x^{-3/4}y^{1/2}$ and $f_y = \frac12 x^{1/4}y^{-1/2}$, so \[ f_x(100, 10) = \frac 14 \times 100^{-3/4} \times 10^{1/2} = \frac{1}{40},\qquad f_y(100, 10) = \frac 12 \times 100^{1/4} \times 10^{-1/2} = \frac{1}{2}. \] and \[ f(100+\Delta x, 10+\Delta y) = 10 + \frac1{40}\Delta x + \frac12\Delta y + \text{error} \] With $\Delta x=1$, $\Delta y = -1$ we get \[ f(101, 9) \approx 10 + \frac 1{40} - \frac 12 . \]
You are moving through the plane along some path $(x(t), y(t))$. Along the way you observe the values of a function $z=f(x,y)$. At some particular time $t_0$ the gradient of the function at your location has length $5$, your speed is $24$, and your velocity vector makes a 60 degree angle with the gradient of $f$. Compute the rate of change in $z$ that you observe at time $t_0$.
(we won't worry about it in this course, but if it bothers you that the length of the gradient and the speed and velocity here have no units, then you could add those units yourself. Assume that the quantity $z=f(x,y)$ is the local temperature, measured in degrees C, length measured in meters, time in seconds. What units does the gradient have?)

$\frac{d\,f(\vec x(t))}{dt} = \vec\nabla f\cdot \vx'(t_0) = \|\vec\nabla f\|\, \|\vx'(t_0)\| \cos \theta = 5\times 24\times \cos 60^\circ = \dots$.
Use implicit differentiation to find the partial derivatives of a function $z=f(x,y)$ that satisfies $xy+xz+z^3=3$
We have $xy+xf(x,y)+f(x,y)^3=3$ for all $(x,y)$. Differentiate both sides wrt $x$ to get \[ y+ f+xf_x + 3f^2f_x = 0. \] Then solve for $f_x$: $f_x(x,y) = \frac{-y-f(x,y)}{x+3f(x,y)^2}$.
The same procedure gives $f_y = - \frac{-x}{x+3f(x,y)^2} $.

The first derivatives you find are $$z_x = -\frac{y+z}{x+3z^2} \text{ and }z_y = -\frac{x}{x+3z^2}$$ where $z=f(x,y)$. To find the second derivatives, you differentiate these expressions, e.g. $$ z_{xy} = \frac{\partial z_x}{\partial y} = \frac{\partial}{\partial y} \left( -\frac{y+z}{x+3z^2}\right) $$ Apply the quotient rule: $$ z_{xy} = -\frac{(x+3z^2) \frac{\partial (y+z)}{\partial y} - \frac{\partial (x+3z^2)}{\partial y} (y+z)} {(x+3z^2)^2} $$ Remember that $z$ is a function of $x$ and $y$, and that we already have its $y$ derivative: $$ z_{xy} = -\frac{(x+3z^2) (1+z_y) - (0+6z z_y) (y+z)} {(x+3z^2)^2} = -\frac{(x+3z^2) (1+z_y) - 6z z_y (y+z)} {(x+3z^2)^2} $$ To complete the calculation you substitute $z_y = -\frac{x}{x+3z^2}$ which leads to this (not very nice) answer: $$ z_{xy} = -\frac{(x+3z^2) (1-\frac{x}{x+3z^2}) +6z \frac{x}{x+3z^2}(y+z)} {(x+3z^2)^2} $$ You could then simplify this, if necessary. You can compute the other derivatives in the same way. The answers, which will be posted tomorrow, are also not very pretty.
Compute the first and second order derivatives of $g(u,v) = f(x(u,v), y(u,v))$ for certain given $x(u,v)$, $y(u,v)$.
See piazza for an example. The relevant question + answers has number 122: to find it, you type "@122" in the piazza search box in the top left corner of the piazza page.
For which values of $a$, $b$, $c$, and $d$ does a function $z=f(x,y)$ exist whose partial derivatives are given by \[ f_x(x,y) = a x^2y^3 + b x^4y + c y^2 + d x^2,\qquad f_y(x,y) = xy - 12 x^3y^2\;\; ??? \] How do you compute $f(x,y)$ if it exists?
Also, see this similar problem.