Midterm 1 Review
Announcements
- Quiz 5: Grade is out; Question 2 is graded as a bonus question.
- Next week: No quiz next week.
-
CalcLab:
- Place: PMA Building, 8th floor (end of the hallway, follow signs for Department of Mathematics)
- Time: Mon–Thu 2:00–7:00 pm, Fri 2:00–5:00 pm
- My time slot: Thursday 5:00–6:00 pm
Topics to prepare
Friday, October 2, during lecture class time.
| Topic | Section |
|---|---|
| Parametric and polar curves | §10.1–10.3 |
| Polar area | §10.4 |
| Ellipses | §10.5 |
| Spheres | §12.1 |
| Unit vectors | §12.2 |
| Dot products, angles, and vector projections | §12.3 |
Parametric curves
$x=f(t)$, $y=g(t)$, $\alpha\le t\le\beta$.
| Task | Method |
|---|---|
| Eliminate $t$ | Solve for $t$, or use an identity such as $\cos^2 u+\sin^2 u=1$. |
| Identify the curve | Put the Cartesian equation in standard form. Keep any restrictions from the parameter interval. |
| Find points | Substitute each value of $t$ into both $x(t)$ and $y(t)$. |
| Find the direction | Start at $(f(\alpha),g(\alpha))$ and follow points in increasing order of $t$. |
| Check repeated tracing | Compare the parameter interval with the period of the pair $(x(t),y(t))$. |
For $A,B>0$, the ellipse has center $(h,k)$ and horizontal and vertical radii $A,B$. Increasing $u$ traces it counterclockwise; a change of $2\pi$ gives one revolution.
Quiz 1, version A: $x=1+2\cos 2t,\ y=-2+3\sin 2t,\ 0\le t\le\frac{3\pi}{2}$.
(a) Eliminate $t$: $\cos2t=(x-1)/2$ and $\sin2t=(y+2)/3$. Add their squares:
(b) Identify: ellipse; center $(1,-2)$; horizontal radius $2$, vertical radius $3$.
(c) Points:
| $t$ | $2t$ | $(x,y)$ |
|---|---|---|
| $0$ | $0$ | $(3,-2)$ |
| $\pi/4$ | $\pi/2$ | $(1,1)$ |
| $\pi/2$ | $\pi$ | $(-1,-2)$ |
| $3\pi/4$ | $3\pi/2$ | $(1,-5)$ |
| $\pi$ | $2\pi$ | $(3,-2)$ |
| $3\pi/2$ | $3\pi$ | $(-1,-2)$ |
(d) Tracing: start at $(3,-2)$ and move counterclockwise. Since $2t$ runs from $0$ to $3\pi$, the ellipse is traced $1\frac12$ times; the upper half is traced twice.
Blue: first revolution. Magenta: retraced upper half.
Calculus with parametric curves
Write $x'=dx/dt$ and $y'=dy/dt$. The derivative formulas below require $x'\ne0$.
| Quantity | Formula | Condition |
|---|---|---|
| Slope | $\dfrac{dy}{dx}=\dfrac{y'}{x'}$ | $x'\ne0$ |
| Tangent at $t=t_0$ | $y-y(t_0)=\dfrac{y'(t_0)}{x'(t_0)}\bigl(x-x(t_0)\bigr)$ | $x'(t_0)\ne0$ |
| Second derivative | $\dfrac{d^2y}{dx^2}=\dfrac{\frac{d}{dt}(y'/x')}{x'}$ | Positive: concave up. Negative: concave down. |
| Area above the $x$-axis | $A=\displaystyle\int_\alpha^\beta y(t)x'(t)\,dt$ | $y\ge0$; trace once from left to right. |
| Arc length | $L=\displaystyle\int_\alpha^\beta\sqrt{x'^2+y'^2}\,dt$ | Trace the arc once. |
| Surface area about the $x$-axis | $S=2\pi\displaystyle\int_\alpha^\beta |y(t)|\sqrt{x'^2+y'^2}\,dt$ | Trace the generating curve once; count each part of the surface once. |
Quiz 2, version A: $x=t^2,\ y=\frac23t^3,\ 1\le t\le2$, so $x'=2t$ and $y'=2t^2$.
1. Point and tangent at $t=1$
$P=(1,\frac23)$ and $\dfrac{dy}{dx}=\dfrac{y'}{x'}=\dfrac{2t^2}{2t}=t$, so the slope at $t=1$ is $1$.
2. Second derivative and concavity
Apply the second derivative formula $\dfrac{d^2y}{dx^2}=\dfrac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$:
Since $\frac{d^2y}{dx^2}>0$, the curve is concave upward throughout the interval.
3. Area under the curve
Apply the area formula $A=\displaystyle\int_\alpha^\beta y(t)x'(t)\,dt$ (since $y>0$ and $x'>0$):
Magenta: tangent of slope $t$. Shading: area from $t=1$.
Polar coordinates and tracing
In $(r,\theta)$, measure $\theta$ from the positive $x$-axis. Move $r$ units along that ray; if $r<0$, move $|r|$ units in the opposite direction.
Example: $(2,\pi/3)$ gives $(1,\sqrt3)$; $(-2,\pi/3)$ gives $(-1,-\sqrt3)$.
To trace $r=f(\theta)$, increase $\theta$, calculate $r$, and plot the point. When $r=0$, the curve passes through the pole.
Check when the curve starts retracing. A full turn of $\theta$ may draw the same curve more than once.
Recognizing a circle in polar form
For $r=2\cos\theta$, multiply both sides by $r$ so that $r^2=x^2+y^2$ and $r\cos\theta=x$ can be used:
The circle has center $(1,0)$ and radius $1$. It passes through the pole and $(2,0)$.
Similarly, $r=2\sin\theta$ gives $x^2+(y-1)^2=1$, a circle centered at $(0,1)$.
Tangent slopes for polar curves
Treat $\theta$ as the parameter. For $r=f(\theta)$, write $r'=dr/d\theta$.
The product rule gives
Example: $r=2\cos\theta$ at $\theta=\pi/4$
Here $r=\sqrt2$ and $r'=-2\sin\theta=-\sqrt2$. The point is $(1,1)$.
Substitute into the two derivative formulas:
The slope is $0/(-2)=0$, so the tangent is $y=1$.
Horizontal: $y'=0$, $x'\ne0$.
Vertical: $x'=0$, $y'\ne0$.
If both vanish, investigate further.
Polar area from sectors
A sector of radius $r$ and angle $\Delta\theta$ has area $\frac12r^2\Delta\theta$ (with the angle in radians).
Approximate a region by narrow sectors and add their areas:
Between two curves, each slice is an outer sector minus an inner sector:
Sweep the region once. Require $0\le r_{\rm in}\le r_{\rm out}$ on the interval; split it if the boundaries switch roles.
Quiz 3: the region and its bounds
Find the area inside $r=2\cos\theta$ and outside $r=1$.
The blue circle has center $(1,0)$ and radius $1$. Exclude the cyan unit disk; the magenta region remains.
1. Find the intersection angles
2. Choose the interval through $\theta=0$
A ray crosses the region where $2\cos\theta\ge1$:
At $\theta=0$, the ray runs from $r=1$ to $r=2$. At both endpoints, the two radii agree.
Quiz 3: set up and evaluate the area
1. Outer radius $2\cos\theta$, inner radius $1$
Subtract $1^2$ to exclude the unit disk from every slice.
2. Use $\cos^2\theta=(1+\cos2\theta)/2$, then integrate
Since $\int\cos2\theta\,d\theta=\frac12\sin2\theta$, keep the factor $\frac12$.
3. Upper value minus lower value
Quiz 3: the other versions
Each version uses a unit circle whose center is one unit from the pole. Rotating the whole region changes the bounds but preserves its area.
| Version | Outer curve | Angular interval |
|---|---|---|
| A | $r=2\cos\theta$ | $[-\pi/3,\pi/3]$ |
| B | $r=2\sin\theta$ | $[\pi/6,5\pi/6]$ |
| C | $r=-2\cos\theta$ | $[2\pi/3,4\pi/3]$ |
| D | $r=-2\sin\theta$ | $[7\pi/6,11\pi/6]$ |
The inner radius is $1$ in every version. Each interval has width $2\pi/3$.
Ellipses
Let $X=x-h$, $Y=y-k$, and $a>b>0$. The center is $(h,k)$.
| Quantity | Horizontal major axis | Vertical major axis |
|---|---|---|
| Standard form | $\dfrac{X^2}{a^2}+\dfrac{Y^2}{b^2}=1$ | $\dfrac{X^2}{b^2}+\dfrac{Y^2}{a^2}=1$ |
| Focal distance | $c=\sqrt{a^2-b^2}$ | $c=\sqrt{a^2-b^2}$ |
| Vertices | $(h\pm a,k)$ | $(h,k\pm a)$ |
| Co-vertices | $(h,k\pm b)$ | $(h\pm b,k)$ |
| Foci | $(h\pm c,k)$ | $(h,k\pm c)$ |
| Axis lengths | Major: $2a$, minor: $2b$ | Major: $2a$, minor: $2b$ |
The larger denominator is $a^2$ and determines the major-axis direction. For a point on the ellipse, the sum of its distances to the two foci is $2a$.
Quiz 4, version A: $9x^2+16y^2-36x+32y-92=0$.
1. Complete the square
Group $x$ and $y$ terms and factor out coefficients:
2. Standard equation
Divide by $144$ so the right side is $1$:
3. Parameters & Key features
| Center | $(h,k) = (2,-1)$ | Orientation | Horizontal ($a^2=16 > b^2=9$) |
| Radii | $a = 4,\quad b = 3$ | Focal distance | $c = \sqrt{16 - 9} = \sqrt{7}$ |
| Vertices | $(2 \pm 4, -1) \;\Longrightarrow\; (-2,-1),\ (6,-1)$ | Foci | $(2 \pm \sqrt{7}, -1)$ |
| Co-vertices | $(2, -1 \pm 3) \;\Longrightarrow\; (2,-4),\ (2,2)$ | ||
Blue: vertices · orange: foci · violet: the triangle relating $a$, $b$, $c$
- Larger denominator ($16$) is under $(x-2)^2$ $\implies$ horizontal major axis.
- For ellipses, $a^2 = b^2 + c^2$, so $c = \sqrt{a^2 - b^2} = \sqrt{7}$.
- To sketch: plot center $(2,-1)$, move $a=4$ left/right for vertices, and move $b=3$ up/down for co-vertices.
Vector formulas
For projection onto $\mathbf a$, assume $\mathbf a\ne\mathbf0$.
| Quantity | Formula | Check |
|---|---|---|
| Distance | $|PQ|=\sqrt{(\Delta x)^2+(\Delta y)^2+(\Delta z)^2}$ | Use coordinate differences |
| Sphere with center $C(h,k,l)$ | $(x-h)^2+(y-k)^2+(z-l)^2=r^2$ | Center $(h,k,l)$, radius $r$ |
| Displacement | $\overrightarrow{AB}=B-A$ | Terminal point minus initial point |
| Unit vector | $\mathbf a/|\mathbf a|$ | $\mathbf a\ne\mathbf0$ |
| Dot product | $\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3$ | Nonzero vectors are perpendicular when the dot product is $0$ |
| Angle | $\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|\,|\mathbf b|}$, $\ 0\le\theta\le\pi$ | $\mathbf a,\mathbf b\ne\mathbf0$ |
| Scalar and vector projections | $\operatorname{comp}_{\mathbf a}\mathbf b=\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|}$, $\ \operatorname{proj}_{\mathbf a}\mathbf b=\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf a|^2}\,\mathbf a$ | Scalar: divide by $|\mathbf a|$. Vector: divide by $|\mathbf a|^2$, then multiply by $\mathbf a$. |
Version A: center $C=(1,-2,3)$, point $P=(3,1,9)$ on the sphere.
1. Find the radius $r$
Displacement vector from center $C$ to point $P$:
Radius is the length $|\overrightarrow{CP}|$, so:
2. Standard equation
Insert center $(h,k,l) = (1,-2,3)$ and $r^2 = 49$:
3. Expanded form
Expand each square and collect all terms on the left:
- Right-hand side: Must be $r^2 = 49$, not $r = 7$.
- Center signs: Center $(1,-2,3) \implies (x-1)^2 + (y+2)^2 + (z-3)^2$.
- Constant term: $h^2 + k^2 + l^2 - r^2 = 1 + 4 + 9 - 49 = -35$.
- Check point $P(3,1,9)$: $9 + 1 + 81 - 6 + 4 - 54 - 35 = 0$ ✓
Version A: $\mathbf u=\langle2,3,6\rangle$, $\mathbf v=\langle1,2,2\rangle$. Then $|\mathbf u|=7$, $|\mathbf v|=3$, and $\mathbf u\cdot\mathbf v=20$.
(a) Unit vector in direction of $\mathbf u$
Divide each component of $\mathbf u$ by its magnitude $|\mathbf u| = \sqrt{4+9+36} = 7$:
(b) Angle $\theta$ between $\mathbf u$ and $\mathbf v$
Use the dot product formula $\mathbf u \cdot \mathbf v = |\mathbf u||\mathbf v|\cos\theta$:
(c) Vector projection of $\mathbf v$ onto $\mathbf u$
Scale $\mathbf u$ by the projection coefficient $\frac{\mathbf u \cdot \mathbf v}{|\mathbf u|^2}$:
Drawn to scale in the plane containing $\mathbf u$ and $\mathbf v$. Click button to swap target.
- Unit vector length: Check that $|\mathbf{\hat u}|^2 = (2/7)^2 + (3/7)^2 + (6/7)^2 = 49/49 = 1$.
- Scalar vs. Vector: $\operatorname{comp}_{\mathbf u}\mathbf v = 20/7$ (scalar length); $\operatorname{proj}_{\mathbf u}\mathbf v$ is a vector.
- Projection onto $\mathbf u$: Denominator is $|\mathbf u|^2 = 49$, and direction is along $\mathbf u$.
For every version, $\mathbf u=P-C$ and $|\mathbf v|=3$.
| Ver. | Sphere | Unit vector | Angle | $\operatorname{proj}_{\mathbf u}\mathbf v$ |
|---|---|---|---|---|
| A$C(1,-2,3)$ $P(3,1,9)$ $\langle1,2,2\rangle$ |
$(x-1)^2+(y+2)^2+(z-3)^2=49$ $x^2+y^2+z^2-2x+4y-6z-35=0$ |
$\langle 2/7,\ 3/7,\ 6/7\rangle$ | $\arccos(20/21)\approx17.8^\circ$ | $(20/49)\langle2,3,6\rangle$ |
| B$C(-2,1,4)$ $P(1,5,4)$ $\langle2,1,2\rangle$ |
$(x+2)^2+(y-1)^2+(z-4)^2=25$ $x^2+y^2+z^2+4x-2y-8z-4=0$ |
$\langle 3/5,\ 4/5,\ 0\rangle$ | $\arccos(2/3)\approx48.2^\circ$ | $\langle 6/5,\ 8/5,\ 0\rangle$ |
| C$C(2,3,-1)$ $P(4,0,5)$ $\langle2,2,1\rangle$ |
$(x-2)^2+(y-3)^2+(z+1)^2=49$ $x^2+y^2+z^2-4x-6y+2z-35=0$ |
$\langle 2/7,\ -3/7,\ 6/7\rangle$ | $\arccos(4/21)\approx79.0^\circ$ | $(4/49)\langle2,-3,6\rangle$ |
| D$C(-1,-2,2)$ $P(3,-2,5)$ $\langle1,2,2\rangle$ |
$(x+1)^2+(y+2)^2+(z-2)^2=25$ $x^2+y^2+z^2+2x+4y-4z-16=0$ |
$\langle 4/5,\ 0,\ 3/5\rangle$ | $\arccos(2/3)\approx48.2^\circ$ | $\langle 8/5,\ 0,\ 6/5\rangle$ |
Keep angles in exact $\arccos$ form. Check that unit vectors have magnitude $1$.
Formula summary
Core formulas and conditions across all Midterm 1 topics.
1. Parametric Curves (§10.1–10.2)
Area: require $y \ge 0$ and trace once from left to right ($x' > 0$).
2. Polar Curves & Area (§10.3–10.4)
Require $r_{\rm out} \ge r_{\rm in} \ge 0$ and sweep the region exactly once.
3. Ellipses (§10.5)
The larger denominator is $a^2$. If under $(y-k)^2$, the major axis is vertical.
4. Spheres & Vectors (§12.1–12.3)
Nonzero vectors: dot product is $0$ $\iff$ perpendicular. Vector projection has direction $\mathbf a$.