Math 214 - Multivariable Calculus - Prep

Task 18.1

In first semester calculus, differential notation is $dy=f' dx$. At $x=c$, the tangent line passes through the point $P=(c,f(c))$. If $Q=(x,y)$ is any other point on the line, then the vector $\vec {PQ} = (x-c,y-f(c))$ tells us that when $dx=x-c$ we have $dy=y-f(c)$. Substitution give us an equation for the tangent line tangent line as $$\underbrace{y-f(c)}_{dy}={f'(c)}\underbrace{(x-c)}_{dx}.$$ This equation tells us that a change in the output ($y-f(c)$) equals the derivative times a change in the input ($x-c$). In this task, we'll repeat this process to obtain an equation of a tangent plane to a function $f(x,y)$, where differential notation gives $$dz = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy.$$

Consider the function $z=f(x,y)=9-x^2-y^2$. We'll be finding an equation of the tangent plane to $f$ at $(x,y)=(2,1)$. Here is surface plot along with the tangent plane at $(2,1,f(2,1))$, together with a contour plot.

Compute the partial derivatives $f_x$ and $f_y$, and the differential $dz$. At the point $(x,y) = (2,1)$, evaluate the partial derivatives and the function $z=f(x,y)$.
One point on the tangent plane to the surface at $(2,1)$ is the point $P=(2,1,f(2,1))$. Let $Q=(x,y,z)$ be another point on this plane. Use the vector $\vec{PQ}$ to obtain $dz$ when $dx = x-2$ and $dy = y-1$.
We'd like an equation of the tangent plane to $f(x,y)$ when $x=2$ and $y=1$. Differential notation tells us that $$\underbrace{z-?}_{dz}=(-4)\underbrace{(x-?)}_{dx}+(?)\underbrace{(y-?)}_{dy}.$$ Fill in the blanks to get an equation of the tangent plane.
Rewrite the equation you got in the form $A(x-a)+B(y-b)+C(z-c)=0$ and state a normal vector to the plane.
The level curve of $f$ that passes through $(2,1)$ has no change in height, so $dz=0$. Use this fact to give an equation of the tangent line to this level curve at $(2,1)$.

Now let $z=f(x,y)=x^2+4xy+y^2$. At the point $P=(x,y)=(3,-1)$, we'll give an equation of the tangent plane to the surface and an equation of the tangent line to the level curve of $f$ that passes through this point.

Give an equation of the tangent plane at $P=(x,y)=(3,-1)$. [Hint: Find $f_x$, $f_y$, $dx$, $dy$, and then $dz$, all at $(x,y)=(3,-1)$. Then substitute, as done above.]
The level curve of $f$ that passes through $P$ is a curve in the plane. Give an equation of the tangent line to this curve at $P$. [Hint: Since we're on a level curve, what does $dz$ equal? The equation is almost identical to the previous part, with one minor change based on what $dz$ equals.]

The tangent plane and tangent line you just found are shown below.

Task 18.2

A rover moves on a hill where elevation is given by $z=f(x,y)=9-x^2-y^2$. The rover's path is parametrized by $\vec r(t)=(2\cos t, 3\sin t)$.

At time $t=0$, what is the rover's position $\vec r(0)$, and what is the elevation $f(\vec r(0))$ at that position? Then find the positions and elevations at $t=\pi/2$, $t=\pi$, and $t=3\pi/2$ as well.
In the plane, draw the rover's path for $t\in [0,2\pi]$. Then, on the same 2D graph, include a contour plot of the elevation function $f$. Include the level curves that pass through the points in part 1. Along each level curve drawn, state the elevation of the curve. [If you end up with an ellipse and several concentric circles, then you've done this right.]
As the rover follows its elliptical path, the elevation is rising and falling. At which $t$ does the elevation reach a maximum? A minimum? Explain, using your graph.
As the rover moves past the point at $t=\pi/4$, is the elevation increasing or decreasing? In other words, is $\dfrac{df}{dt}$ positive or negative? Use your graph to explain.

Notice above that we wanted $\frac{df}{dt}$, the rate of change of elevation with respect to time, even though the function $f(x,y)$ does not explicitly have $t$ as an input. The proper notation would be $\frac{d(f\circ r)}{dt}$, but this is so cumbersome that it's generally avoided. The notation $\frac{df}{dt}$ requires the reader to infer from context that $x$ and $y$ depend on $t$.

At the point $\vec r(t)$, we'd like a formula for the elevation $f(\vec r(t))$. What is the elevation of the rover at any time $t$? [In $f(x,y)$, replace $x$ and $y$ with what they are in terms of $t$.]
Compute $df/dt$ (the derivative as you did in first-semester calculus).

Let's repeat the above, but first compute differentials before substitution. For reference, we let $f(x,y)=9-x^2-y^2$ and $(x,y)=\vec r(t)=(2\cos t, 3\sin t)$.

Compute the differential $df$ in terms of $x$, $y$, $dx$, and $dy$.
Compute $dx$ and $dy$ in terms of $t$ and $dt$.
Use substitution to write $df$ in terms of $t$ and $dt$. Then divide by $dt$ to obtain $\frac{df}{dt}$. Did you get the same answer as the previous part?
Use your work above to state both $\vec\nabla f(x,y)$ and $\frac{d\vec r}{dt}$. Show that $\frac{df}{dt} = \vec\nabla f(x,y)\cdot \frac{d\vec r}{dt}$.

Task 18.3

A second-order partial derivative of $f$ is a partial derivative of one of the partial derivatives of $f$. The second-order partial of $f$ with respect to $x$ and then $y$ is the quantity $\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial x}\right]$, so we first compute the partial of $f$ with respect to $x$, and then compute the partial of the result with respect to $y$. Alternate notations exist, for example the same second-order partial above we can write as $$\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial x}\right]=\left(f_{x}\right)_y=f_{xy}=\ds\frac{\partial}{\partial y}\frac{\partial}{\partial x}f = \frac{\partial}{\partial y}\frac{\partial f}{\partial x} = \frac{\partial^2 f}{\partial y \partial x}.$$ The subscript notation $f_{xy}$ is easiest to write. Sometimes we'll use subscript notation to mean something other than a partial derivative (like the $x$ or $y$ component of a vector), at which point we use the fractional partial derivative notation to avoid confusion.

Consider the functions $f(x,y,z) = xy^2z^3$ and $g(x,y)=x\cos(xy)$.

First compute $\vec \nabla f$. Then compute $f_{xy}$ and $\frac{\partial^2 f}{\partial z\partial y}$. Explain how you got these. End by computing the entire second derivative $D\vec\nabla f(x,y,z)$ (it is a 3 by 3 matrix with all 9 second partials placed inside).
Compute $g_x$ and then $g_{xy}$. Then compute $g_y$ followed by $g_{yx}$.
Now let $f(x,y)=3xy^3+e^{x}.$ Compute the four second partials $$\ds \frac{\partial^2 f}{ \partial x^2},\quad \ds\frac{\partial^2 f}{\partial y \partial x},\quad \ds\frac{\partial^2 f}{\partial y^2}, \quad \text{ and }\ds\frac{\partial^2 f}{\partial x \partial y}.$$
For $f(x,y)=x^2\sin(y)+y^3$, compute both $f_{xy}$ and $f_{yx}$.
Make a conjecture about a relationship between $f_{xy}$ and $f_{yx}$. Then use your conjecture to quickly compute $f_{xy}$ if $$f(x,y)=3xy^2+\tan^{2}(\cos(x)) (x^{49}+x)^{1000}.$$

Task 18.4

The last problem for prep each day will point to relevant problems from OpenStax. Spend 30 minutes working on problems from the sections below.

Return to any of the previous day's OpenStax problems to locate extra practice.

Big View
Prep Day 18 Task 18.1 In first semester calculus, differential notation is $dy=f' dx$. At $x=c$, the tangent line passes through the point $P=(c,f(c))$. If $Q=(x,y)$ is any other point on the line, then the vector $\vec {PQ} = (x-c,y-f(c))$ tells us that when $dx=x-c$ we have $dy=y-f(c)$. Substitution give us an equation for the tangent line tangent line as $$\underbrace{y-f(c)}_{dy}={f'(c)}\underbrace{(x-c)}_{dx}.$$ This equation tells us that a change in the output ($y-f(c)$) equals the derivative times a change in the input ($x-c$). In this task, we'll repeat this process to obtain an equation of a tangent plane to a function $f(x,y)$, where differential notation gives $$dz = \frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy.$$ Consider the function $z=f(x,y)=9-x^2-y^2$. We'll be finding an equation of the tangent plane to $f$ at $(x,y)=(2,1)$. Here is surface plot along with the tangent plane at $(2,1,f(2,1))$, together with a contour plot. Compute the partial derivatives $f_x$ and $f_y$, and the differential $dz$. At the point $(x,y) = (2,1)$, evaluate the partial derivatives and the function $z=f(x,y)$. One point on the tangent plane to the surface at $(2,1)$ is the point $P=(2,1,f(2,1))$. Let $Q=(x,y,z)$ be another point on this plane. Use the vector $\vec{PQ}$ to obtain $dz$ when $dx = x-2$ and $dy = y-1$. We'd like an equation of the tangent plane to $f(x,y)$ when $x=2$ and $y=1$. Differential notation tells us that $$\underbrace{z-?}_{dz}=(-4)\underbrace{(x-?)}_{dx}+(?)\underbrace{(y-?)}_{dy}.$$ Fill in the blanks to get an equation of the tangent plane. Rewrite the equation you got in the form $A(x-a)+B(y-b)+C(z-c)=0$ and state a normal vector to the plane. The level curve of $f$ that passes through $(2,1)$ has no change in height, so $dz=0$. Use this fact to give an equation of the tangent line to this level curve at $(2,1)$. Now let $z=f(x,y)=x^2+4xy+y^2$. At the point $P=(x,y)=(3,-1)$, we'll give an equation of the tangent plane to the surface and an equation of the tangent line to the level curve of $f$ that passes through this point. Give an equation of the tangent plane at $P=(x,y)=(3,-1)$. [Hint: Find $f_x$, $f_y$, $dx$, $dy$, and then $dz$, all at $(x,y)=(3,-1)$. Then substitute, as done above.] The level curve of $f$ that passes through $P$ is a curve in the plane. Give an equation of the tangent line to this curve at $P$. [Hint: Since we're on a level curve, what does $dz$ equal? The equation is almost identical to the previous part, with one minor change based on what $dz$ equals.] The tangent plane and tangent line you just found are shown below. Task 18.2 A rover moves on a hill where elevation is given by $z=f(x,y)=9-x^2-y^2$. The rover's path is parametrized by $\vec r(t)=(2\cos t, 3\sin t)$. At time $t=0$, what is the rover's position $\vec r(0)$, and what is the elevation $f(\vec r(0))$ at that position? Then find the positions and elevations at $t=\pi/2$, $t=\pi$, and $t=3\pi/2$ as well. In the plane, draw the rover's path for $t\in [0,2\pi]$. Then, on the same 2D graph, include a contour plot of the elevation function $f$. Include the level curves that pass through the points in part 1. Along each level curve drawn, state the elevation of the curve. [If you end up with an ellipse and several concentric circles, then you've done this right.] As the rover follows its elliptical path, the elevation is rising and falling. At which $t$ does the elevation reach a maximum? A minimum? Explain, using your graph. As the rover moves past the point at $t=\pi/4$, is the elevation increasing or decreasing? In other words, is $\dfrac{df}{dt}$ positive or negative? Use your graph to explain. Notice above that we wanted $\frac{df}{dt}$, the rate of change of elevation with respect to time, even though the function $f(x,y)$ does not explicitly have $t$ as an input. The proper notation would be $\frac{d(f\circ r)}{dt}$, but this is so cumbersome that it's generally avoided. The notation $\frac{df}{dt}$ requires the reader to infer from context that $x$ and $y$ depend on $t$. At the point $\vec r(t)$, we'd like a formula for the elevation $f(\vec r(t))$. What is the elevation of the rover at any time $t$? [In $f(x,y)$, replace $x$ and $y$ with what they are in terms of $t$.] Compute $df/dt$ (the derivative as you did in first-semester calculus). Let's repeat the above, but first compute differentials before substitution. For reference, we let $f(x,y)=9-x^2-y^2$ and $(x,y)=\vec r(t)=(2\cos t, 3\sin t)$. Compute the differential $df$ in terms of $x$, $y$, $dx$, and $dy$. Compute $dx$ and $dy$ in terms of $t$ and $dt$. Use substitution to write $df$ in terms of $t$ and $dt$. Then divide by $dt$ to obtain $\frac{df}{dt}$. Did you get the same answer as the previous part? Use your work above to state both $\vec\nabla f(x,y)$ and $\frac{d\vec r}{dt}$. Show that $\frac{df}{dt} = \vec\nabla f(x,y)\cdot \frac{d\vec r}{dt}$. Task 18.3 A second-order partial derivative of $f$ is a partial derivative of one of the partial derivatives of $f$. The second-order partial of $f$ with respect to $x$ and then $y$ is the quantity $\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial x}\right]$, so we first compute the partial of $f$ with respect to $x$, and then compute the partial of the result with respect to $y$. Alternate notations exist, for example the same second-order partial above we can write as $$\frac{\partial}{\partial y}\left[\frac{\partial f}{\partial x}\right]=\left(f_{x}\right)_y=f_{xy}=\ds\frac{\partial}{\partial y}\frac{\partial}{\partial x}f = \frac{\partial}{\partial y}\frac{\partial f}{\partial x} = \frac{\partial^2 f}{\partial y \partial x}.$$ The subscript notation $f_{xy}$ is easiest to write. Sometimes we'll use subscript notation to mean something other than a partial derivative (like the $x$ or $y$ component of a vector), at which point we use the fractional partial derivative notation to avoid confusion. Consider the functions $f(x,y,z) = xy^2z^3$ and $g(x,y)=x\cos(xy)$. First compute $\vec \nabla f$. Then compute $f_{xy}$ and $\frac{\partial^2 f}{\partial z\partial y}$. Explain how you got these. End by computing the entire second derivative $D\vec\nabla f(x,y,z)$ (it is a 3 by 3 matrix with all 9 second partials placed inside). Compute $g_x$ and then $g_{xy}$. Then compute $g_y$ followed by $g_{yx}$. Now let $f(x,y)=3xy^3+e^{x}.$ Compute the four second partials $$\ds \frac{\partial^2 f}{ \partial x^2},\quad \ds\frac{\partial^2 f}{\partial y \partial x},\quad \ds\frac{\partial^2 f}{\partial y^2}, \quad \text{ and }\ds\frac{\partial^2 f}{\partial x \partial y}.$$ For $f(x,y)=x^2\sin(y)+y^3$, compute both $f_{xy}$ and $f_{yx}$. Make a conjecture about a relationship between $f_{xy}$ and $f_{yx}$. Then use your conjecture to quickly compute $f_{xy}$ if $$f(x,y)=3xy^2+\tan^{2}(\cos(x)) (x^{49}+x)^{1000}.$$ Task 18.4 The last problem for prep each day will point to relevant problems from OpenStax. Spend 30 minutes working on problems from the sections below. Return to any of the previous day's OpenStax problems to locate extra practice. Edit Page History Source Attach File Backlinks List Group Page last modified on October 10, 2022, at 03:38 PM
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Day 18

Task 18.1

Task 18.2

Task 18.3

Task 18.4