I-Learn, Class Pictures, Learning Targets, Text Book Practice
Prep Tasks: Unit 1 - Motion, Unit 2 - Derivatives, Unit 3 - Integration, Unit 4 - Vector Calculus

Day 11 - Prep

Task 11.1

We'll use Mathematica to create a program to compute all the quantities related to the TNB frame and curvature.

Download the Mathematica file TNB-intro.nb.
Read through the introductory examples, evaluating each block of code.
Adapt the code to compute the TNB frame and curvature for a given curve.

Task 11.2

Recall that the curvature vector is $\kappa = \dfrac{d\vec T}{ds}$, with curvature being a the length of this vector. This vector tells us how much the direction of motion ($\vec T$) changes, as we increase the distance moved along the curve. As such, a tight corner will result in a large change of direction and hence a large curvature. Large corners will result in a small curvature.

The radius of curvature at a point, namely $1/\kappa$, provides the radius of a circle that approximates the shape of curve at that point. Large turns results in a large radius of curvature, while tight turns results in a small radius of curvature. This circle lies in the plane formed by $\vec T$ and $\vec N$ (so a normal vector to this plane is $\vec B$). We call this plane the osculating plane. The center of the circle can be found by following $\vec N$ from the point on the curve.

For the curve $\vec r(t) = (3\cos t, 3\sin t, 4t)$, we have already computed $\vec T$, $\vec N$, $\vec B$, and $\kappa$. At $t=\pi/2$, evaluate these quantities.
Give an equation of the the osculating plane at $t=\pi/2$. You'll need to identify a normal vector to the plane, and a point on the plane.
Explain why the center of curvature is given by $\vec r + \frac{1}{\kappa}\vec N$.
Give the location of the center of curvature for $\vec r(t) = (3\cos t, 3\sin t, 4t)$ at $t=\pi/2$.

Task 11.3

There are many ways to compute the TNB frame and curvature. In this problem, we'll develop a few others.

Explain why $ \vec N = \dfrac{\vec r^{\prime\prime}_{\perp \vec r^{\prime}}}{ |\vec r^{\prime\prime}_{\perp \vec r^{\prime} }| } $.
Explain why $ \vec B = \dfrac{\vec r^{\prime}\times \vec r^{\prime\prime}}{|\vec r^{\prime}\times \vec r^{\prime\prime}|}$.
For a function of the form $\vec r(x) = (x, f(x))$, show that $\kappa = \dfrac{|f''(x)|}{(1+(f'(x))^2)^{3/2}}$.

Task 11.4

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

section 3.3 exercises 113-151

Day 11 - In class

Brain Gains (Rapid Recall, Jivin' Generation)

A curve passes through the point $P = (x,y,z)$ and has normal vector $(a,b,c)$ with curvature $\kappa$ at that point. State the center of curvatuve.
Give an equation of a plane that passes through the points $P = (1,2,3)$, $(-1,3,0)$, and $(0,4,2)$.
For the function $V = \pi r^2 h$, assume that both $r$ and $h$ depend on a parameter $t$. Then compute $\frac{dV}{dt}$ using implicit differentiation.

Group Problems

For the function $z = x^2y+y^2$, compute $\frac{dz}{dt}$ (assuming that both $x$ and $y$ depend on the parameter $t$).
The computations for the TNB frame for the curve $r(t) = (t,0,t^2)$ get rather ugly really quickly, if we leave everything in terms of $t$. At a single point, we can simplify things, provided we evaluate AFTER all differentiation has occurred. Compute $\vec T(0)$, $\vec N(0)$, $\vec B(0)$, and $\kappa(0)$, as well as state the center of curvature at $t=0$.
Pick an exercise from section 3.3, exercises 113-151. Tackle it together.

Day 12 - Prep

Task 12.1

A computer chip has been disconnected from electricity and sitting in cold storage for quite some time. The chip is connected to power, and a few moments later the temperature (in Celsius) at various points $(x,y)$ on the chip is measured. From these measurements, statistics is used to create a temperature function $z=f(x,y)$ to model the temperature at any point on the chip. Suppose that this chip's temperature function is given by the equation $z=f(x,y)=9-x^2-y^2$. (This could just as easily have been the elevation of a rover at a point $(x,y)$ on a hill.) We'll be creating both a 2D contour plot (topographical map) and 3D surface plot of this function in this task.

The points in the plane with temperature $f(x,y)=0$ satisfy $0=9-x^2-y^2$, or equivalently $x^2+y^2=9$. These points lie on a circle of radius 3, so we can draw that circle in the $xy$-plane (the start of our 2D contour plot) and also in 3D by plotting a circle of radius 3 at height $z=0$ (the start of our 3D surface plot). These two plots are shown below.

What is the temperature at $(0,0)$, $(1,2)$, and $(-4,3)$?
Which points in the plane have temperature $z=5$? Add this contour (level curve) to your 2D contour plot. Then at height $z=5$, add the same curve to the 3D surface plot.
Repeat the above for $z=8$, $z=9$, and $z=1$. What's wrong with letting $z=10$?
Letting $y = 0$ provides a vertical cross section of the surface. This is the curve $z = 9-x^2-0^2$. This curve cannot be drawn on the contour plot, but can be added to your 3D surface plot. Add that curve, and then add the curve given by letting $x=0$.
Describe the 3D surface that you created with your plot. Add any extra features to your 3D surface plot to convey the 3D image you constructed. You can use the Mathematica file ContourSurfaceGradient.nb to check your work.
For the function $f(x,y) = x^2-y$, construct a 2D contour plot and 3D surface plot.

Task 12.2

Suppose that an explosion occurs at the origin $(0,0,0)$. Heat from the explosion starts to radiate outwards. Suppose that a few moments after the explosion, the temperature at any point in space is given by $w=T(x,y,z)=100-x^2-y^2-z^2.$

Which points in space have a temperature of 99? To answer this, replace $T(x,y,z)$ by $99$ to get $99=100-x^2-y^2-z^2$. Use algebra to simplify this to $x^2+y^2+z^2=1$. Draw this object.
Which points in space have a temperature of 96? of 84? Draw the surfaces.
What is the temperature at $(3,0,-4)$? Draw the set of points that have this same temperature.
The 4 surfaces you drew above are called level surfaces. If you walk along a level surface, what happens to your temperature?
When we compute a level surface of a function $w = f(x,y,z)$, which variable do we make constant? When we compute a level curve of a function $z=f(x,y)$, which variable do we make constant?
Consider now the function $w=f(x,y,z)=x^2+z^2$. This function has an input $y$, but notice that changing the input $y$ does not change the output of the function.
- Draw a graph of the level surface $w=4$. [When $y=0$ you can draw one curve. When $y=1$, you draw the same curve. When $y=2$, again you draw the same curve. This kind of graph we call a cylinder, and is important in manufacturing where you extrude an object through a hole.]
- Graph the level surface $9=x^2+z^2$ (so $w=9$), and $w=16$.

You can use the Mathematica file ContourSurfaceGradient.nb to check your work.

Task 12.3

Suppose the elevation $z$ of terrain near a rover is given by the formula $z=f(x,y) = x^2+3xy$.

Suppose that $x$ and $y$ are both functions of $t$, and then use implicit differentiation to compute $\dfrac{dz}{dt}$. Write your answer in the form $$\frac{dz}{dt} = (?)\frac{dx}{dt}+(?)\frac{dy}{dt}.$$
The differential of $z$ (or differential of $f$ as $z=f(x,y)$) is obtained by multiplying both sides above by $dt$. Verify that $dz = (2x+3y)dx+3xdy$.
Write the differential of $f$ as the dot product $$df = (?,?)\cdot(dx,dy).$$

When we write the differential of a function $f(x,y)$ in the form $df = M dx +N dy$, we call $M$ the partial derivative of $f$ with respect to $x$, written $f_x$ or $\frac{\partial f}{\partial x}$ or $D_x f$, and we call $N$ the partial derivative of $f$ with respect to $y$, written $f_y$ or $\frac{\partial f}{\partial y}$ or $D_y f$. The vector $(f_x,f_y)$ we call the gradient of $f$, written as $\vec\nabla f$, which means the differential of $f$ is always $$df = \vec \nabla f \cdot (dx,dy) = (f_x, f_y)\cdot (dx,dy) = f_xdx+f_ydy.$$ Similar definitions hold for functions of more variables.

For the function $f(x,y)=3x^2+2xy$, compute the differential $df$ (in terms of $x$, $y$, $dx$, $dy$), the partial derivatives $f_x$ and $f_y$, and the gradient $\vec \nabla f(x,y)$.
For the function $g(r,s,t)=r^2s^3+4rt^2$ compute the differential $dg$ (in terms of $r$, $s$, $t$, $dr$, $ds$, $dt$), the partial derivatives $g_r$ and $\frac{\partial g}{\partial s}$ and $D_tg$, and the gradient $\vec \nabla g(r,s,t)$.

Task 12.4