Math 234 — Study guide 2nd midterm

Topics covered on the second midterm

Linear functions

1. What is a linear function?
2. The graph of a linear function is a plane: how do you find a normal vector to this plane?

Quadratic forms

1. What is a quadratic form?
2. How do you classify a quadratic form?
3. If a form is indefinite, then what is its zero set?

Partial derivatives

The linear approximation formula

1. What does it say?
2. How do you use it to approximate (for example) $f(2.01, 1.97)$ for some given function, if you know $f$ and its derivatives $f_x$ and $f_y$ at $(2,1)$?

Tangent planes to the graph of a function

1. Given a function $z=f(x,y)$ how do you find the tangent plane to the graph at some given point $(a,b,f(a,b))$ on the graph?
2. How do you find the normal vector to the tangent plane?

Chain rule

1. How do you differentiate $f(x(t), y(t))$ with respect to $t$?
2. If $x$ and $y$ each are functions of two variables $u$ and $v$ instead of just one (e.g. $t$), then how do you differentiate $g(u,v) = f(x(u,v), y(u,v))$ with respect to $u$?
3. Or with respect to $v$? See problems 1 — 5, Chapter 4, §12, page 76.

Gradients

1. What is the gradient of a function $z=f(x, y)$? Or of a function $w=F(x, y, z)$?
2. How do you write the linear approximation formula using vectors?
3. If $\vx(t) = \vek x(t)\\ y(t)\tor$ is a vector function, then how do you write the chain rule for $\frac d{dt} f(x(t), y(t))$ using vectors?
4. True or false: the gradient of $f(x, y)$ at $(a,b)$ is a normal vector to the tangent plane to the graph of $z=f(x, y)$ at the point $(a,b, f(a,b))$ ???
5. How do you find the tangent plane to the level set of a function of three variables $w=F(x, y, z)$ at a given point $(a,b,c)$?

Implicit functions

1. If a function $y=f(x)$ of one variable satisfies an equation $F(x, y)=c$ for some constant, then how do you write $f'(x)$ in terms of the partial derivatives of $F$?
2. What does the Implicit function theorem say?

Brandon Alberts’ practice problems

Some more typical problems

1. 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 $F_x(x_0,y_0,z_0)(x-x_0) + F_y(x_0,y_0,z_0) (y-y_0) + F_z(x_0,y_0,z_0) (z-z_0) = 0$
2. 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 \[ f(101, 9) \approx 10 + \frac 1{40} - \frac 12 . \]
3. 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 $t_0$.
   $\frac{d\,f(\vec x(t))}{dt} = \vec\nabla f\cdot \vx'(t_0) = \|\vec\nabla f\|\, \|\vx'(t_0)\| \cos \theta = 5\times 24\times \cos 60^\circ = \dots$.
4. 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 $z=f(x,y)$. For the second midterm you will only be asked to find the first derivatives.

   If this problem appears on the final exam, then you could also be asked to find one or all of the second derivatives. To do this, 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.

5. Compute the partial derivatives of $g(u,v) = f(x(u,v), y(u,v))$ for certain given $x(u,v)$, $y(u,v)$.