MIT 18.01 Problem Set 8A

This lesson reproduces the Problem Set “MIT 18.01 Problem Set 8A” from MIT OpenCourseWare 18.01 Single Variable Calculus (Fall 2006) (https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/), licensed under CC BY-NC-SA 4.0. Work through 3 problems in order: type your reasoning in the answer box or write it on the handwriting pad, then submit for AI grading.

How the grading works Every problem is graded by an AI mathematics teacher that reads the official solution shipped with the course. It checks your work criterion by criterion, awards partial credit for correct steps, and returns written feedback. Equivalent mathematics and valid alternative methods are accepted — a different route to the same result can still earn full marks.

If your answer is incomplete, the feedback names exactly which steps are missing, so you can correct them and try again.

(Lec 30, 7 pts: $3+1+3$)

a) Find the algebraic equation in $x$ and $y$ for the curve

Draw the portion of the curve $0 \leq t \leq \pi/2$ in the three cases $k=1$, $k=2$, $k=3$.

b) Without calculation, find the arclength in the cases $k=1$ and $k=2$.

c) Find a definite integral formula for the length of the curve for general $k$. Then evaluate the integral in the three cases $k=1$, $k=2$, and $k=3$. (Your answer in the first two cases should match what you found in part (b), but the calculation takes more time.)

(Lec 30, 9 pts: $3+1+3+2$) The hyperbolic sine and cosine are defined by

a) Show that

i) $\dfrac{d}{dx}\sinh x=\cosh x$ and $\dfrac{d}{dx}\cosh x=\sinh x$,

ii) $\cosh^2 x=1+\sinh^2 x$ and

iii) $\cosh^2 x=\dfrac{1+\cosh 2x}{2}$

b) What curve is described parametrically by $x=\cosh t$, $y=\sinh t$? (Give the equation and its name.)

c) The curve $y=\cosh x$ is known as a \emph{catenary}. It is the curve formed by a chain whose two ends are held at the same height.

i) Sketch the curve

ii) Find its arclength from the lowest point to the point $(x_1, \cosh x_1)$ for a fixed $x_1 > 0$

d) Find the area of the surface of revolution formed by revolving the portion of the curve from part (c) around the $x$-axis. This surface is known as a \emph{catenoid}. It is interesting because it is the surface of least area connecting the two circles that form its edges. If you dip two circles of wire in a soap solution, then (with some coaxing) a soap film will form in this shape. In general, the soap films try to span a frame of wires with a surface with the least area possible.

(Lec 31, 11 pts: $3+2+2+4$) Area in polar coordinates.

a) Find the area of the right triangle with vertices at $(x,y)=(0,0)$, $(x,y)=(a,0)$ and at $(x,y)=(a,h)$, using polar coordinates. (One of the less convenient ways to find the area of a triangle.)

b) Find the equation in $(x,y)$ coordinates for the curve $r=1/(1+\sin\theta)$ and sketch it.

c) Find the area of the region $0 \leq r \leq 1/(1+\sin\theta)$, $0 \leq \theta \leq \pi$, using the \emph{rectangular} coordinate formula you found in part (b).

d) Find the area of the region in part (c) using polar coordinates. One way to evaluate the integral is to change variables to $u=\theta-\pi/2$, and then use the half angle formula

This area was already computed in part (c), but the polar coordinate formula is still valuable because it gives the area of any sector, not just the one that is bounded by a horizontal line. The area swept out from the viewpoint of the focus of an ellipse, parabola, or hyperbola is related by Kepler's law to the speed of planets and comets.