Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)
Pearson eText for Calculus for the Life Sciences -- Instant Access (Pearson+)
2nd Edition
ISBN: 9780137553457
Author: Raymond Greenwell, Nathan Ritchey
Publisher: PEARSON+
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Chapter P, Problem 22PSDT

Simplify as a single term without negative exponents: k 1 m 1 .

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This question builds on an earlier problem. The randomized numbers may have changed, but have your work for the previous problem available to help with this one. A 4-centimeter rod is attached at one end to a point A rotating counterclockwise on a wheel of radius 2 cm. The other end B is free to move back and forth along a horizontal bar that goes through the center of the wheel. At time t=0 the rod is situated as in the diagram at the left below. The wheel rotates counterclockwise at 1.5 rev/sec. At some point, the rod will be tangent to the circle as shown in the third picture. A B A B at some instant, the piston will be tangent to the circle (a) Express the x and y coordinates of point A as functions of t: x= 2 cos(3πt) and y= 2 sin(3t) (b) Write a formula for the slope of the tangent line to the circle at the point A at time t seconds: -cot(3πt) sin(3лt) (c) Express the x-coordinate of the right end of the rod at point B as a function of t: 2 cos(3πt) +411- 4 -2 sin (3лt) (d)…
5. [-/1 Points] DETAILS MY NOTES SESSCALCET2 6.5.AE.003. y y= ex² 0 Video Example x EXAMPLE 3 (a) Use the Midpoint Rule with n = 10 to approximate the integral कर L'ex² dx. (b) Give an upper bound for the error involved in this approximation. SOLUTION 8+2 1 L'ex² d (a) Since a = 0, b = 1, and n = 10, the Midpoint Rule gives the following. (Round your answer to six decimal places.) dx Ax[f(0.05) + f(0.15) + ... + f(0.85) + f(0.95)] 0.1 [0.0025 +0.0225 + + e0.0625 + 0.1225 e0.3025 + e0.4225 + e0.2025 + + e0.5625 €0.7225 +0.9025] The figure illustrates this approximation. (b) Since f(x) = ex², we have f'(x) = 0 ≤ f'(x) = < 6e. ASK YOUR TEACHER and f'(x) = Also, since 0 ≤ x ≤ 1 we have x² ≤ and so Taking K = 6e, a = 0, b = 1, and n = 10 in the error estimate, we see that an upper bound for the error is as follows. (Round your final answer to five decimal places.) 6e(1)3 e 24( = ≈
2. [-/1 Points] DETAILS MY NOTES SESSCALCET2 6.5.015. Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n. (Round your answers to six decimal places.) ASK YOUR TEACHER 3 1 3 + dy, n = 6 (a) the Trapezoidal Rule (b) the Midpoint Rule (c) Simpson's Rule Need Help? Read It Watch It
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