Calc 3 Midterm Prep

PROBLEM 1: Spheres in 3D

Q1: Where does the sphere equation come from? Why does it have that form?

• A sphere is the set of all points that are the same distance $r$ from a center $(h,k,l)$.

• By the 3D distance formula (the Pythagorean theorem applied twice), the distance from $(x,y,z)$ to the center is:$\sqrt{(x-h)^2+(y-k)^2+(z-l)^2}$

• Setting that equal to $r$ and squaring both sides gives:$(x-h)^2+(y-k)^2+(z-l)^2=r^2$

PROBLEM 1: Spheres in 3D

How did you find the radius?

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

• The sphere passes through $(6,-3,3)$, so that point is exactly $r$ from the center $(-3,6,3)$.

• Any point on a sphere is exactly $r$ from the center, so one point plus the center is enough.

• I used the distance formula between those two points. Since the equation has $r^2$ on the right side, I could just use r² directly: $r^2=9^2+(-9)^2+0^2=162$

PROBLEM 1: Spheres in 3D:

Why did you set $z=0$?

What is this curve called?

• The $xy$-plane is every point where $z=0$.

• A point on the intersection has to be on both the sphere and the plane, so it has to satisfy both equations at once.

• Substituting $z=0$ into the sphere's equation does exactly that.

• The resulting curve is called a TRACE.

PROBLEM 1: Spheres in 3D:

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

What does that tell you about the center?

• The $z$-term becomes $(0-3)^2=9$ because the center has $z$-coordinate 3, so it's 3 units above the $xy$-plane.

• That 9 is the squared distance from the center to the plane.

• Subtracting it from $r^2=162$ gives the circle's squared radius, 153.

• It's the Pythagorean theorem in a cross-section: the hypotenuse is the sphere's radius $\sqrt{162}$, one leg is the height 3, and the other leg is the circle's radius $\sqrt{153}$.

• That's why the circle is smaller than the sphere. If the center were in the plane, the term would vanish and the circle would have the full radius $\sqrt{162}$.

PROBLEM 1: Spheres in 3D:

Does every sphere intersect the $xy$-plane?

How would you know from the equation?

No. After substituting $z=0$, the right side is $r^2-l^2$.

• $r^2-l^2>0$: a CIRCLE (the plane cuts through).

• $r^2-l^2=0$: a SINGLE POINT (the plane just touches).

• $r^2-l^2<0$: NO INTERSECTION, since a sum of squares can't be negative. Example: with the center at $z=20$ we'd get $(x+3)^2+(y-6)^2=162-400=-238$, which is impossible. Here $162-9=153>0$, so it's a circle.

PROBLEM 1: Spheres in 3D:

How do you know your answer is correct?

How could you check it?

• For (a), plug the given point $(6,-3,3)$ into the equation: $(6+3)^2+(-3-6)^2+(3-3)^2=81+81+0=162$

• That matches $r^2$ (not $r$), so the point is on the sphere.

• For (b), the circle's $\rho^2=153$ should equal $r^2-l^2=162-9$, and it does.

• I could also take a point on the circle, like $(-3+\sqrt{153},\,6,\,0)$, and confirm it satisfies the sphere's equation.

PROBLEM 2: Work and the Dot Product:

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

• Only the part of the force in the direction of motion does work.

• That part has magnitude $|\mathbf{F}|\cos\theta$, so the work is $|\mathbf{F}|\cos\theta\cdot|\mathbf{D}|$.

• That's exactly the geometric definition of the dot product, $\mathbf{F}\cdot\mathbf{D}=|\mathbf{F}||\mathbf{D}|\cos\theta$.

• In components it's $F_1D_1+F_2D_2+F_3D_3$, so I don't need to find $\theta$.

PROBLEM 2: Work and the Dot Product:

How did you get $\mathbf{D}$?

What would happen to $W$ if you subtracted in the wrong order?

• I subtracted the starting point from the end point in each coordinate: $\mathbf{D}=\langle 5,6,6\rangle$.

• Subtracting the other way flips every component's sign, so $\mathbf{D}$ would become $\langle -5,-6,-6\rangle$and it would point along the same line but in the opposite direction, and $W$ would come out as $-77$ J.

• That would wrongly say the force works against the motion.

PROBLEM 2: Work and the Dot Product:

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

Would that give a bigger or smaller answer?

• The force and displacement don't point the same way, and only the part of the force along the displacement does work.

• Multiplying magnitudes assumes they're parallel, which overstates the work:

• $|\mathbf{F}|=\sqrt{75}\approx 8.66$ N,

• $|\mathbf{D}|=\sqrt{97}\approx 9.85$ m, so $|\mathbf{F}||\mathbf{D}|=\sqrt{7275}\approx 85.3$ J

  • which is more than 77 J.

  • The dot product can never be larger than $|\mathbf{F}||\mathbf{D}|$. They're equal only when the force and displacement point the same direction.

PROBLEM 2: Work and the Dot Product:

What is the angle between $\mathbf{F}$ and $\mathbf{D}$?

What does the positive answer tell you?

$\cos\theta=\frac{\mathbf{F}\cdot\mathbf{D}}{|\mathbf{F}||\mathbf{D}|}=\frac{77}{\sqrt{75}\sqrt{97}}=\frac{77}{\sqrt{7275}}\approx 0.9028$, so $\theta\approx 25.5^\circ$. Since $\theta<90^\circ$, $\cos\theta>0$, so the work is positive: the force mostly helps the motion. About 90% of the force points along the displacement, which is why 77 J is about 90% of 85.3 J.

• $\theta<90^\circ$: $W>0$ (helps the motion)

• $\theta=90^\circ$: $W=0$ (perpendicular, does nothing)

• $\theta>90^\circ$: $W<0$ (opposes the motion)

PROBLEM 2: Work and the Dot Product:

Why are the units joules?

• The force is in newtons and the displacement is in meters.

• Each term in the dot product, like $F_1D_1$, is newtons times meters, so the work is in newton-meters.

• One newton-meter is defined as one joule: $1\text{ J}=1\text{ N}\cdot\text{m}$.

PROBLEM 2: Work and the Dot Product:

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

What if the force weren't constant?

• For a CONSTANT force, only the start and end points matter. Chop any path into small pieces $\Delta\mathbf{D}_i$

  • Then:$\sum\mathbf{F}\cdot\Delta\mathbf{D}_i=\mathbf{F}\cdot\sum\Delta\mathbf{D}_i=\mathbf{F}\cdot\mathbf{D}$, because $\mathbf{F}$ is the same on every piece and the pieces add up to the total displacement.

• So a curved path would also give 77 J. The "straight line" just makes it clear that $\mathbf{D}$ is the displacement.

• If the force VARIED along the path, $\mathbf{F}$ couldn't be pulled out of the sum. I'd need to know the path and use a line integral, $W=\int_C\mathbf{F}\cdot d\mathbf{r}$ (later in Calc 3).

PROBLEM 2: Work and the Dot Product:

What is the projection of one vector onto another?

How does it relate to work?

• SCALAR PROJECTION (a number): $\text{comp}_{\mathbf{a}}\mathbf{b}=|\mathbf{b}|\cos\theta=\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|}$

• VECTOR PROJECTION (a vector): $\text{proj}_{\mathbf{a}}\mathbf{b}=\left(\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}|^2}\right)\mathbf{a}$

• GEOMETRICALLY: the shadow of $\mathbf{b}$ on the line through $\mathbf{a}$ (drop a perpendicular from the tip of $\mathbf{b}$). The leftover $\mathbf{b}-\text{proj}_{\mathbf{a}}\mathbf{b}$ is perpendicular to $\mathbf{a}$. The scalar projection is negative when $\theta>90^\circ$.

• WORK: the component of the force along the motion times the distance. $\text{comp}_{\mathbf{D}}\mathbf{F}=\frac{77}{\sqrt{97}}\approx 7.82$ N, so $W=\frac{77}{\sqrt{97}}\cdot\sqrt{97}=77$ J. Of the 8.66 N force, 7.82 N acts along the motion; the perpendicular rest does no work.

  • Vector version: $\text{proj}_{\mathbf{D}}\mathbf{F}=\frac{77}{97}\langle 5,6,6\rangle\approx\langle 3.97,4.76,4.76\rangle$.

• Link to Problem 4: a dot product of 0 means each projection is zero, i.e. perpendicular.

PROBLEM 2: Work and the Dot Product:

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

Proof with the LAW OF COSINES:

Put $\mathbf{a}$ and $\mathbf{b}$ tail to tail with angle $\theta$; the third side is $\mathbf{a}-\mathbf{b}$.

1. Law of Cosines: $|\mathbf{a}-\mathbf{b}|^2=|\mathbf{a}|^2+|\mathbf{b}|^2-2|\mathbf{a}||\mathbf{b}|\cos\theta$

2. Dot product: $|\mathbf{a}-\mathbf{b}|^2=(\mathbf{a}-\mathbf{b})\cdot(\mathbf{a}-\mathbf{b})=|\mathbf{a}|^2-2\,\mathbf{a}\cdot\mathbf{b}+|\mathbf{b}|^2$

3. Set equal: the squared lengths cancel, leaving $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$

PROBLEM 3: The Normal Plane:

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

• The normal plane at a point is the plane through that point that's perpendicular to the curve's direction of motion.

• A plane is determined by a point and a normal vector, meaning a vector perpendicular to the plane.

• The curve's direction is given by the tangent vector $\mathbf{r}'(t)$. Since the plane is perpendicular to that direction, $\mathbf{r}'(t)$ is perpendicular to the plane, which is exactly what a normal vector is.

• (The normal plane contains every line through the point that's perpendicular to the tangent.)

PROBLEM 3: The Normal Plane:

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

• A plane's normal vector determines its ORIENTATION, meaning how it's tilted in space.

  • Every plane perpendicular to a given direction has the same tilt.

    • Two planes are parallel when they have the same tilt, so they must be perpendicular to the same direction.

      • That means their normal vectors point along the same line (the same way or opposite ways), so the normals are parallel.

• Picture a stack of paper sheets with a pencil standing straight up through them.

PROBLEM 3: The Normal Plane:

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

• Parallel planes need parallel normal vectors, not equal ones. Two vectors are parallel when one is a SCALAR MULTIPLE of the other, so I set $\mathbf{r}'(t)=k\langle 3,3,-4\rangle$.

• The length of a normal vector doesn't matter, only its direction. Here $k=1$, but I couldn't assume that ahead of time.

• Example: the same plane written as $6x+6y-8z=4$ has normal $\langle 6,6,-8\rangle$.

  • Setting $\mathbf{r}'(t)$ equal to that would give $3=6$ (no solution), while using $k$ gives $k=\tfrac{1}{2}$ and the same $t=-1$.

PROBLEM 3: The Normal Plane:

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

• $t=1$ satisfies the $x$-component (since $3t^2=3$ gives $t=\pm 1$)

  • but it fails the $z$-component: $4(1)^3=4$, not $-4$.

  • At $t=1$, $\mathbf{r}'(1)=\langle 3,3,4\rangle$, which isn't a scalar multiple of $\langle 3,3,-4\rangle$.

• Only $t=-1$ satisfies all three components at once.

PROBLEM 3: The Normal Plane:

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

• Both planes have the same coefficients, $\langle 3,3,-4\rangle$, so they have the same normal vector and the same tilt.

• The normal plane is $3(x+1)+3(y+3)-4(z-1)=0$, or $3x+3y-4z=-16$, so the constants differ ($-16$ vs $2$).

• More convincingly, plug the point $(-1,-3,1)$ into the given plane: $3(-1)+3(-3)-4(1)=-16\neq 2$.

  • The point isn't on the given plane, so they're two separate parallel planes.

• "Different equations" alone isn't enough, since $6x+6y-8z=4$ is the same plane as $3x+3y-4z=2$

PROBLEM 4: Angle Between Two Space Curves:

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

• Curves don't have a single direction, so the angle between them is DEFINED at the intersection point using their tangent lines.

• Close up, each curve looks like its tangent line

• The tangent vectors $\mathbf{r}_1'$ and $\mathbf{r}_2'$ point along those lines, so the angle between the curves is the angle between the tangent vectors, which I find with the dot product formula $\cos\theta=\frac{\mathbf{r}_1'\cdot\mathbf{r}_2'}{|\mathbf{r}_1'||\mathbf{r}_2'|}$.

PROBLEM 4: Angle Between Two Space Curves:

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

Why check each curve separately?

• I set each component equal to the matching coordinate of $(2,0,0)$.

• I solved the easiest component first and checked that the value works in the other two, since $t$ must satisfy all three at once.

• I did this for each curve separately because each has its own parameter, and they could reach the point at different $t$-values.

PROBLEM 4: Angle Between Two Space Curves:

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

• I need the angle between the TANGENT vectors, and $\mathbf{r}'(t)$ is the tangent vector: it points in the direction the curve is moving.

• $\mathbf{r}(t)$ is the POSITION vector, pointing from the origin to the point.

• At the intersection, both curves have the same position, $\langle 2,0,0\rangle$, so using $\mathbf{r}(t)$ would give $0^\circ$ for any two curves crossing there.

  • That says nothing about how the curves are actually moving.

PROBLEM 4: Angle Between Two Space Curves:

What does a dot product of 0 tell you?

• Since $\mathbf{a}\cdot\mathbf{b}=|\mathbf{a}||\mathbf{b}|\cos\theta$ and neither vector is zero, a dot product of 0 forces $\cos\theta=0$, so $\theta=90^\circ$.

• The vectors are perpendicular: neither has any component in the other's direction.

  • $\mathbf{a}\cdot\mathbf{b}>0$ : acute angle (partly the same way)

  • $\mathbf{a}\cdot\mathbf{b}=0$: $90^\circ$ (perpendicular)

  • $\mathbf{a}\cdot\mathbf{b}<0$: obtuse angle (partly opposite ways)

PROBLEM 4: Angle Between Two Space Curves:

What if the formula gave an obtuse angle?

• Two crossing lines make two angles, $\theta$ and $180^\circ-\theta$.

• Which one the formula gives depends on the direction each curve is traced, since reversing a parameter ($t\to -t$) flips its tangent vector.

• By convention, the angle between curves is the acute one, so I'd report $180^\circ-\theta$, or use $\cos\theta=\frac{|\mathbf{r}_1'\cdot\mathbf{r}_2'|}{|\mathbf{r}_1'||\mathbf{r}_2'|}$. Here the angle is $90^\circ$, so both choices agree.

PROBLEM 5: Tangent Line to a Parametric Curve:

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, the rate of change of $y$ with respect to $x$

• Slope is rise over run, and $\frac{dy}{dt}$ and $\frac{dx}{dt}$ are the rates of rise and run as $t$ changes

• Formally, it comes from the CHAIN RULE: $\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}$, so dividing by $\frac{dx}{dt}$ gives $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

PROBLEM 5: Tangent Line to a Parametric Curve:

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

What if $\frac{dx}{dt}=0$?

• The formula divides by $\frac{dx}{dt}$, so it needs $\frac{dx}{dt}\neq 0$

• Geometrically, $\frac{dx}{dt}=0$ means there's no horizontal motion at that instant

  • If $\frac{dy}{dt}\neq 0$, the curve is moving straight up or down, so the tangent is VERTICAL with undefined slope

PROBLEM 5: Tangent Line to a Parametric Curve:

Why did you need the product rule?

Both $x$ and $y$ are PRODUCTS OF TWO FUNCTIONS OF $t$:

• $4t\cdot\cos t$ and

• $$4t\cdot\sin t$$

  • Since both factors change with $t$, I need the product rule

PROBLEM 5: Tangent Line to a Parametric Curve:

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

I plugged $t=-\pi$ into both equations:

• $x=4(-\pi)(-1)=4\pi$ and

• $$y=4(-\pi)(0)=0$$

  • So the point is $(4\pi,0)$

PROBLEM 5: Tangent Line to a Parametric Curve Q5: How can you check that your answer makes sense?

1. The point is on the line: $y=-\pi(4\pi)+4\pi^2=0$.
2. The direction makes sense: the velocity $\langle -4,4\pi\rangle$ means the curve moves left slowly and up quickly (about $\pi$ times faster). Up-and-left gives a negative slope, and the ratio gives $|m|\approx 3$. A slope of $-\pi$ fits both. (This check catches a flipped fraction like $-1/\pi$, which would be nearly flat.)
3. It matches the graph: the tangent at $(4\pi,0)$ is steep and slopes downward. I also made sure to use radians, since the trig functions in calculus are in radians.

PROBLEM 6: Polar Equation of a Conic Q1: Why did you rewrite the equation first?

The standard forms all have a 1 as the constant in the denominator. That's what makes the coefficient of $\sin\theta$ equal to $e$. The original, $-\frac{7}{2\sin\theta-1}$, has $-1$ as the constant and an extra minus sign out front, so I couldn't read $e$ or the directrix's side directly. Multiplying the top and bottom by $-1$ gives $\frac{7}{1-2\sin\theta}$, which matches $\frac{ed}{1-e\sin\theta}$, so I can read off $e=2$ and $ed=7$. Example: for $r=\frac{7}{3-6\sin\theta}$, reading $e=6$ would be wrong. Divide the top and bottom by 3 to get $\frac{7/3}{1-2\sin\theta}$, so $e=2$.

PROBLEM 6: Polar Equation of a Conic Q2: How did you get $e$ and $d$, and how do you know where the directrix is?

Matching $\frac{7}{1-2\sin\theta}$ to $\frac{ed}{1-e\sin\theta}$: the coefficient of $\sin\theta$ is $e$, so $e=2$. The numerator is $ed=7$, so $d=\frac{7}{2}$. Because the form has $\sin\theta$, the directrix is HORIZONTAL, a line $y=$ constant. Because the denominator has a MINUS sign, it's BELOW the pole. So the directrix is $y=-\frac{7}{2}$. ($d$ is a distance, so it's always positive. The sign in the denominator gives the side.)

PROBLEM 6: Polar Equation of a Conic Q3: Why does $e$ decide the type of conic? What does $e=2$ mean geometrically?

MEANING OF $e=2$: $e$ is the fixed ratio $\frac{|PF|}{|P\ell|}$. With $e=2$, every point on the curve is TWICE AS FAR from the focus as it is from the directrix. WHY $e$ DECIDES THE SHAPE: look at whether the denominator $1-e\sin\theta$ can be zero (where $r$ blows up to infinity). Since $-1\le\sin\theta\le 1$:

• $e<1$: never zero, so $r$ stays finite and the curve closes: ELLIPSE.
• $e=1$: zero at one angle ($\sin\theta=1$), so it goes to infinity in one direction: PARABOLA.
• $e>1$: zero at two angles ($\sin\theta=\frac{1}{e}$), so it goes to infinity in two directions: HYPERBOLA. Here $e=2$ gives $\sin\theta=\frac{1}{2}$, so $\theta=\frac{\pi}{6},\frac{5\pi}{6}$: exactly the asymptote directions. INTUITION: if $e<1$, points must stay closer to the focus than to the directrix, so the curve hugs the focus and closes. If $e>1$, points can be farther from the focus than from the directrix, so the curve escapes to infinity, with branches on both sides of the directrix.

PROBLEM 6: Polar Equation of a Conic Q4: Why is $r$ negative at $\theta=\frac{\pi}{2}$, and what does that mean for the sketch?

At $\theta=\frac{\pi}{2}$, $\sin\theta=1$, so the denominator is $1-2=-1$ and $r=-7$. A negative $r$ means you go in the OPPOSITE direction from $\theta$. The angle $\frac{\pi}{2}$ points straight up, so $r=-7$ puts the point 7 units straight DOWN, at $(0,-7)$, not $(0,7)$. More generally, whenever $\sin\theta>\frac{1}{2}$ (that is, $\frac{\pi}{6}<\theta<\frac{5\pi}{6}$), the denominator is negative and $r<0$. Those angles trace the entire LOWER BRANCH. All other angles give $r>0$ and trace the upper branch.

PROBLEM 6: Polar Equation of a Conic Q5: Where are the focus, vertices, and asymptotes?

The FOCUS is at the pole, $(0,0)$, because the polar form assumes that. The VERTICES lie on the axis through the focus perpendicular to the directrix, here the $y$-axis. Plugging in $\theta=\frac{3\pi}{2}$ and $\frac{\pi}{2}$ gives $(0,-\frac{7}{3})$ and $(0,-7)$. The CENTER is their midpoint, $(0,-\frac{14}{3})$. The ASYMPTOTES go through the center in the directions where the denominator is zero, $\sin\theta=\frac{1}{2}$, so $\theta=\frac{\pi}{6},\frac{5\pi}{6}$, with slopes $\pm\frac{1}{\sqrt{3}}$: $y=-\frac{14}{3}\pm\frac{x}{\sqrt{3}}$. Check: $a=\frac{7}{3}$ (center to vertex), $c=\frac{14}{3}$ (center to focus), so $e=\frac{c}{a}=2$, which matches.

PROBLEM 6: Polar Equation of a Conic Q6: Where does the polar form $r=\frac{ed}{1-e\sin\theta}$ come from?

Put the focus at the pole and the directrix at $y=-d$. Take a point $P=(r\cos\theta,\ r\sin\theta)$.

1. Distance to the focus: $|PF|=r$.
2. Distance to the directrix (vertical gap to $y=-d$): $|P\ell|=r\sin\theta+d$.
3. Definition: $r=e(d+r\sin\theta)$.
4. Solve: $r-er\sin\theta=ed$, so $r(1-e\sin\theta)=ed$, so $r=\frac{ed}{1-e\sin\theta}$. A directrix above the pole ($y=d$) gives distance $d-r\sin\theta$, which leads to $1+e\sin\theta$. A vertical directrix uses $x=r\cos\theta$, which gives cosine. That's WHY sine means a horizontal directrix and the sign gives the side: it falls out of the distance calculation.

PROBLEM 6: Polar Equation of a Conic Q7: Could you have solved it by converting to Cartesian coordinates?

Yes. It's a good way to VERIFY the answer, though the polar form gives $e$ and the directrix directly, which is why I used it.

1. Clear the fraction: $r-2r\sin\theta=7$.
2. Substitute $y=r\sin\theta$: $r=7+2y$.
3. Square, use $r^2=x^2+y^2$: $x^2+y^2=49+28y+4y^2$, so $x^2-3y^2-28y=49$.
4. Complete the square: $x^2-3\left(y+\frac{14}{3}\right)^2=-\frac{49}{3}$.
5. Standard form: $\frac{\left(y+\frac{14}{3}\right)^2}{49/9}-\frac{x^2}{49/3}=1$. A VERTICAL hyperbola, center $(0,-\frac{14}{3})$, $a=\frac{7}{3}$, $b=\frac{7}{\sqrt{3}}$, $c=\frac{14}{3}$. Vertices $(0,-\frac{7}{3})$ and $(0,-7)$; foci $(0,0)$ and $(0,-\frac{28}{3})$; $e=\frac{c}{a}=2$; asymptotes $y+\frac{14}{3}=\pm\frac{x}{\sqrt{3}}$. All match. Subtle point: squaring adds no extra points. $r<0$ exactly when $y<-\frac{7}{2}$: the lower bran