Using R, provide the solution for any exercise in either Chapter 4 or Chapter 7 of the calculus textbook. If you are unsure of your solution, post your concerns.

Chapter 4 Page 190 Exercise 15

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Hand Solution

Volume Formula \[V = \frac{1}{3}\pi r^2 h\] The height is 2/3 of the diameter. \[h = \frac{2}{3}d\] Switching the height and diameter, the equation of d should be: \[d = \frac{3}{2}h\] The radius is 1/2 of diameter \[r = \frac{1}{2}d\] \[\frac{1}{2}d = \frac{\frac{3}{2}h}{2} \] \[r = 3h\] By plugging in the radius to the volume formula, it should be: \[V = \frac{1}{3}\pi (3h)^2 h\] \[V = \frac{1}{3}\pi 9h^3\] \[V = 3\pi h^3\] The related rate should be: \[\frac{dv}{dt} = 3\pi 3h^2 \frac{dh}{dt}\] By plugging in the rate and the height, the equation and the answer should be: \[5 = 3\pi 3(30)^2 \frac{dh}{dt}\] \[5 = 8100\pi \frac{dh}{dt}\] \[\frac{dh}{dt} = \frac{5}{8100\pi}\] \[\frac{dh}{dt} = 1.96487584\times10^{-4}\]