Calculus I, Fall 2021

Brief solutions to Quiz 8

Dec 21, 2021

1. Find the length of the curve y = f(x) = ∫_{0}^{x} \sqrt{cos t} dt from x = 0 to x = \frac{π}{2}. Ans: f′(x) = \sqrt{cos x}. ds = \sqrt{1 + (f′(x))^{2}}dx.

L = ∫0^{\frac{π}{2}} ds = ∫0^{\frac{π}{2}} \sqrt{1 + (f′(x))^{2}}dx = ∫0^{\frac{π}{2}} \sqrt{1 + cos x} dx = ∫0^{\frac{π}{2}} \sqrt{2 cos^{2} \frac{x}{2}} dx.

= ∫0^{\frac{π}{2}} \sqrt{2} cos \frac{x}{2} dx = 2\sqrt{2} sin \frac{x}{2}|0^{\frac{π}{2}} = 2.

2. A torus (donut) is generated by revolving the circle (x − 2)^{2} + y^{2} = 1 around the y axis. Find the surface area of the torus. \underline{Give all details}. Ans: See page 6 of Lecture 22 note and page 1-2 of Lecture 23 note.

3. Find the solutions for \frac{dy}{dx} = 3x^{2}e^{−y} on \{x ≥ 0\} with y(0) = 0.

Ans:

\frac{dy}{dx} = 3x^{2}e^{−y},

e^{y}dy = 3x^{2}dx, ∫ e^{y}dy = ∫ 3x^{2}dx,

e^{y} = x^{3} + C, y = ln(x^{3} + C), y(0) = 0, therefore C = 1. Answer: y = ln(x^{3} + 1).