Calc 3 Midterm (conversational)

PROBLEM 1:

Where does the sphere equation come from?

Why does it have that form?

A sphere is just all the points that are the same distance $r$ from the center. So I write the 3D distance formula from a point $(x,y,z)$ to the center, set it equal to $r$, and square both sides. Which gives the sphere equation.$(x-h)^2+(y-k)^2+(z-l)^2=r^2$.

PROBLEM 1:

How did you find the radius?

Why can you use $r^2$ directly instead of $r$?

The sphere goes through the point $(6,-3,3)$, so that point is exactly one radius away from the center. I used the distance formula between the two points. Since the equation needs $r^2$ anyway, I didn't bother taking the square root: $r^2=81+81+0=162$.

PROBLEM 1:

Why did you set $z=0$?

What is this curve called?

The $xy$-plane is where $z=0$. Any point on the intersection has to satisfy both the sphere's equation and $z=0$, so I just plug $z=0$ in. The curve you get is called a trace.

PROBLEM 1:

Why is the $z$-term equal to 9 instead of disappearing?

What does that tell you about the center?

The center sits at $z=3$, so it's 3 units above the plane. Plugging in $z=0$ leaves $(0-3)^2=9$, which is that height squared. Taking it away from $162$ leaves $153$, the circle's radius squared. It's really just the Pythagorean theorem in a cross-section, which is why the circle is smaller than the sphere.

PROBLEM 1:

Does every sphere intersect the $xy$-plane?

How would you know from the equation?

No.

• After plugging in $z=0$, the right side is $r^2$ minus the center's height squared.

  • If that's positive, you get a circle.

  • If it's zero, the plane just touches the sphere at one point.

  • If it's negative, there's no intersection, because a sum of squares can't be negative.

• Here it's $162-9=153$, so it's a circle.

PROBLEM 1:

How do you know your answer is correct? How could you check it?

I'd plug the given point back in: $(6+3)^2+(-3-6)^2+0=162$, which matches $r^2$, so the point really is on the sphere. For part (b), I'd check that $153=162-9$.

PROBLEM 2:

Why is work the dot product of $\mathbf{F}$ and $\mathbf{D}$?

Only the part of the force that points along the motion actually does any work. That part is $|\mathbf{F}|\cos\theta$, and multiplying it by the distance gives $|\mathbf{F}||\mathbf{D}|\cos\theta$, which is exactly the dot product $\mathbf{F}\cdot\mathbf{D}$. The nice thing is the component formula means I never have to find the angle.

PROBLEM 2:

How did you get $\mathbf{D}$? What would happen to $W$ if you subtracted in the wrong order?

It's end point minus start point, so $\mathbf{D}=\langle 5,6,6\rangle$. If I subtracted the other way, the vector would point backwards and I'd get $-77$ J, which would say the force is working against the motion.

PROBLEM 2:

Why not just multiply $|\mathbf{F}|$ by the distance?

Would that give a bigger or smaller answer?

Because the force and the motion don't point in the same direction. Multiplying the magnitudes assumes they do, so it overestimates.

PROBLEM 2:

What is the angle between $\mathbf{F}$ and $\mathbf{D}$? What does the positive answer tell you?

Using the dot product formula, $\cos\theta=\frac{77}{\sqrt{75}\sqrt{97}}$, so the angle is about $25.5^\circ$. That's less than $90^\circ$, so the work is positive, which means the force is mostly helping the motion.

PROBLEM 2:

Why are the units joules?

Force is in newtons and distance is in meters, so work comes out in newton-meters, and one newton-meter is a joule.

PROBLEM 2:

Does it matter that the object moves in a straight line?

What if the force weren't constant?

No.

• For a constant force, the work only depends on the start and end points.

• If you split any path into small pieces, $\mathbf{F}$ is the same on every piece, so it factors out and the pieces add up to the total displacement $\mathbf{D}$.

• If the force varied along the path, I'd need a line integral.

PROBLEM 2:

What is the projection of one vector onto another?

How does it relate to work?

• The projection of $\mathbf{F}$ onto $\mathbf{D}$ is how much of the force points along the motion

  • Its length is $\frac{\mathbf{F}\cdot\mathbf{D}}{|\mathbf{D}|}$

    • Work is that amount times the distance, so it comes out to $\mathbf{F}\cdot\mathbf{D}$

• The part of the force that's perpendicular does no work

PROBLEM 2:

Why is $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$?

• It comes from the Law of Cosines.

  • If you put $\mathbf{a}$ and $\mathbf{b}$ tail to tail, the third side of the triangle is $\mathbf{a}-\mathbf{b}$

    • Writing its length squared with the Law of Cosines, and again with the dot product, then setting them equal, everything cancels except $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$

PROBLEM 3:

What is the normal plane, and why is $\mathbf{r}'(t)$ its normal vector?

• The normal plane is the plane that cuts straight across the curve, perpendicular to the direction it's moving

• The tangent vector $\mathbf{r}'(t)$ points in that direction, so it's perpendicular to the plane, and that's what a normal vector is

PROBLEM 3:

Why does "parallel planes" mean you compare normal vectors?

• A plane's normal vector sets its tilt.

• Parallel planes have the same tilt, so their normal vectors have to point along the same line.

PROBLEM 3:

Why set $\mathbf{r}'(t)=k\langle 3,3,-4\rangle$ instead of $\mathbf{r}'(t)=\langle 3,3,-4\rangle$?

• Parallel vectors don't have to be equal, one just has to be a scalar multiple of the other

  • Only the direction of a normal vector matters, not its length

PROBLEM 3:

Why isn't $t=1$ a solution?

• It works for the $x$-component, because squaring hides the sign.

  • But the $z$-component needs $4t^3=-4$, and $t=1$ gives $+4$.

    • So only $t=-1$ works in all three components at once.

PROBLEM 3:

How do you know the two planes are parallel and not the same plane?

• They have the same normal vector, so they're parallel. But if I plug my point $(-1,-3,1)$ into the given plane, I get $-16$ instead of $2$.

• So the point isn't on that plane, which means they're two separate parallel planes.

PROBLEM 4:

Why is the angle between the curves the angle between their tangent vectors?

• Curves don't have one fixed direction, so the angle between them is defined using their tangent lines at the crossing point.

• If you zoom in, each curve looks like its tangent line there.

  • The tangent vectors point along those lines, so I find the angle between them.

PROBLEM 4:

How did you find the $t$-value for each curve?

Why check each curve separately?

I set each component equal to the point's coordinates, solved the easiest one, and then checked that $t$ works in the other two.

• I did each curve separately because they each have their own parameter.

  • Both turned out to be $t=0$, but that's just a coincidence.

PROBLEM 4:

Why did you use $\mathbf{r}'(t)$ and not $\mathbf{r}(t)$?

• $\mathbf{r}'(t)$ is the tangent vector, the direction the curve is moving. $\mathbf{r}(t)$ is just the position.

• At the crossing point, both curves have the same position, so using $\mathbf{r}(t)$ would always give $0^\circ$, which tells you nothing.

PROBLEM 4:

What does a dot product of 0 tell you?

It means the vectors are perpendicular.

• In this problem since $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$ and neither vector is zero, so $\cos\theta$ has to be zero, so the angle is $90^\circ$

PROBLEM 4:

What if the formula gave an obtuse angle?

By convention you report the acute angle, so I'd subtract it from $180^\circ$

PROBLEM 5:

What does $\frac{dy}{dx}$ represent, and where does $\frac{dy/dt}{dx/dt}$ come from?

$\frac{dy}{dx}$ is the slope of the tangent line

• Slope is rise over run, and $\frac{dy}{dt}$ and $\frac{dx}{dt}$ are how fast you're rising and running as $t$ changes.

• Formally, it comes from the chain rule: $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$, and you just divide.

PROBLEM 5:

Why does the formula need $\frac{dx}{dt}\neq 0$? What if $\frac{dx}{dt}=0$?

• Because you'd be dividing by zero.

• Geometrically, $\frac{dx}{dt}=0$ means the curve isn't moving sideways at that instant, so the tangent is vertical and the slope is undefined.

PROBLEM 5:

Why did you need the product rule?

• Both $x$ and $y$ are a product of two functions of $t$

• So the product rule is the only way to differentiate them.

PROBLEM 5:

How do you know $(4\pi,0)$ is the right point?

Where is it on the spiral?

• I just plugged $t=-\pi$ into both equations. $\cos(-\pi)=-1$ and $\sin(-\pi)=0$, so I get $(4\pi,0)$

• It's on the positive $x$-axis, about $12.6$ units out from the origin.

PROBLEM 5:

How can you check that your answer makes sense?

• First, the point satisfies the line's equation.

• Second, the velocity $\langle -4,4\pi\rangle$ says the curve is moving left slowly and up fast, so the slope should be negative and steep, and $-\pi$ is exactly that.

• It also matches the graph.

PROBLEM 6:

Why did you rewrite the equation first?

The standard forms all start with $1$ in the denominator, and that's what makes the coefficient of $\sin\theta$ equal to $e$.

PROBLEM 6:

How did you get $e$ and $d$, and how do you know where the directrix is?

• Comparing with $\frac{ed}{1-e\sin\theta}$: the number in front of $\sin\theta$ is $e$, so $e=2$.

• The top is $ed=7$, so $d=\frac{7}{2}$.

• Sine tells me the directrix is horizontal, and the minus sign tells me it's below the pole, so it's $y=-\frac{7}{2}$.

PROBLEM 6:

Why does $e$ decide the type of conic? What does $e=2$ mean geometrically?

• $e$ is the ratio of the distance to the focus over the distance to the directrix.

  • So $e=2$ means every point is twice as far from the focus as from the directrix.

• The type of conic depends on whether the denominator can hit zero

  • If $e<1$ it never does, so the curve closes up into an ellipse.

  • If $e=1$ it happens at one angle, which is a parabola.

  • If $e>1$ it happens at two angles, so the curve shoots off in two directions, which is a hyperbola.

PROBLEM 6:

Why is $r$ negative at $\theta=\frac{\pi}{2}$, and what does that mean for the sketch?

• At $\frac{\pi}{2}$, $\sin\theta=1$, so the denominator is $1-2=-1$ and $r=-7$.

  • A negative $r$ means you go the opposite way.

  • So instead of 7 units up, the point is 7 units down, at $(0,-7)$

• All the angles between $\frac{\pi}{6}$ and $\frac{5\pi}{6}$ give negative $r$, and that's what traces the lower branch.

PROBLEM 6:

Where are the focus, vertices, and asymptotes?

• The focus is at the origin, because that's built into the polar form.

• The vertices come from plugging in $\frac{\pi}{2}$ and $\frac{3\pi}{2}$: $(0,-7)$ and $(0,-\frac{7}{3})$.

• The center is halfway between them, at $(0,-\frac{14}{3})$.

• The asymptotes go through the center in the directions where the denominator is zero, which gives slopes of $\pm\frac{1}{\sqrt{3}}$.

PROBLEM 6: Where does the polar form $r=\frac{ed}{1-e\sin\theta}$ come from?

• It comes straight from the focus-directrix definition.

• If you put the focus at the origin and the directrix at $y=-d$ , the distance to the focus is just $r$, and the distance to the directrix is $d+r\sin\theta$.

• Setting $r=e(d+r\sin\theta)$ and solving for $r$ gives $\frac{ed}{1-e\sin\theta}$

PROBLEM 6:

Could you have solved it by converting to Cartesian coordinates?

Yes.

• Using $y=r\sin\theta$, the equation becomes $r=7+2y$.

  • Squaring it and completing the square gives a standard hyperbola centered at $(0,-\frac{14}{3})$ with $a=\frac{7}{3}$ and $c=\frac{14}{3}$. So $e=\frac{c}{a}=2$, which matches.

  • I used the polar form because it gives $e$ and the directrix direct

PROBEM 6:

How would you convert it to cartesian coordinates?

1. Clear the fraction.

2. Replace $r\sin\theta$ with $y$ (or $r\cos\theta$ with $x$).

3. Square. Replace $r^2$ with $x^2 + y^2$.

4. Complete the square to get a standard form.