University:
Whitman College
Course:
MATH-225 | Calculus III
Academic year:

2023
Views:

368

Pages:

4
Author:

Selena Curtis

KEY Math 225: Quiz the First 9.6.06 This exam is closed book and closed notes. You may use your calculator for the purposes of arithmetic and for plotting equations, if helpful. When asked for specific values, however, you must show the relevant algebra. You have until 5 minutes before the hour to finish. 1. Suppose (x(t), y(t)) are a set of parametric equations for a ≤ t ≤ b. To reverse the direction of the parametric curve, what algebraic operation do we need to perform? Replace & with (t) in both equations and limits. (d) 2. What is the relationship between a circle's radius to a point and its tangent line to that same point? Radius is perpendicular to tangent. 3. What is the difference between the graph of the line y1 and the parametric curves (cos(t), 1-cos(t)) y=1-x full line, all xé y valves (cos(t), (1-cost)) only values -14x51 and o≤ y ≤2. 4. Give the formula for the slope of a tangent line to a polar curver (8). dy= rcost+d sme -r sint cost dy dx 1Title: Match the following equations (x(t), y(t)) to their respective graphs. Give reasons for your choices. Note: There is one extra group! Equations: (a) ((\sin(t), \sin(t) \cos(t))) (b) ((\ln(1-t), 1)) (c) ((\sin(t), 1-\sin(t))) (d) ((\sin(4t), \sin(5t))) (e) ((\sin(4t), \sin(5t))) Graph Labels: I, II, III, IV, V, VI Annotations: (a) bounded in y, only goes through origin (d) bounded in x, y intercepts twice per period (e) x positive only, y positive & negative; both unbounded (b) y between 0 and 1 only (c) y between 0 and 2 only Title: (b) Find the total area enclosed by this curve. Note: You should only have to do one integral. How does your answer depend on n? Equations: [\int_{0}^{\frac{\pi}{n}}\frac{1}{4}(\sin(6\theta))^{2}d\theta] [=\int_{0}^{\frac{\pi}{n}}\frac{1}{4}(1-\cos(2n\theta))d\theta] [=\frac{1}{4}\left[\theta-\frac{\sin(2n\theta)}{2n}\right]_{0}^{\frac{\pi}{n}}] [=\frac{\pi}{4n}-\frac{1}{4n}(0-0)] [=\frac{\pi}{4n}] n even + 2n leaves total area = [2n \cdot \frac{\pi}{4n} = \frac{\pi}{2}] n odd + n leaves total area = [n \cdot \frac{\pi}{4n} = \frac{\pi}{4}] Annotation: Depends only on parity, not on value. Graph Description: (Bonus) See the graphs below. They represent the graph of the polar equations (x = \sin(t)) and (y = \cos(2t)), which we showed in class to be a parabola similar to (y = 1 - 2x^2), periodic. The three graphs have increasing bounds on t. Why might we get different pictures for the graphs?

Calculus III Group Work, Section 10.1

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