Question
The slope of a function f (x) at any point (x, y) is \(\frac{x-3}{x^{2}-3x-4}\) The point \(\left ( 5,\frac{4}{5}\ln 6 \right )\) is on the graph of f (x).
(A) Write an equation of the tangent line to the graph of f (x) at x = 5.
(B) Use the tangent line in part (a) to approximate f (4.5) to the nearest thousandth.
(C) Find the antiderivative of \(\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{x-3}{x^{2}-3x-4}\) with the condition \(f(5)=\frac{4}{5}\ln 6\).
(D) Use the result of part (c) to find f (4.5) to the nearest thousandth.
Answer/Explanation
Question
Grass clippings are placed in a bin, where they decompose. For 0 ≤ t ≤ 30 , the amount of grass clippings remaining in the bin is modeled by A(t) = 6.687 (0.931)t , where A(t) is measured in pounds and t is measured in days.
(a) Find the average rate of change of A(t) over the interval 0 ≤ t ≤ 30. Indicate units of measure.
(b) Find the value of A'(15) . Using correct units, interpret the meaning of the value in the context of the problem.
(c) Find the time t for which the amount of grass clippings in the bin is equal to the average amount of grass clippings in the bin over the interval 0 ≤ t ≤ 30 .
(d) For t >30, L (t), the linear approximation to A at t =30, is a better model for the amount of grass clippings remaining in the bin. Use L (t) to predict the time at which there will be 0.5 pound of grass clippings remaining in the bin. Show the work that leads to your answer.
Answer/Explanation
Ans:
(a) \(\frac{A(30)-A(0)}{30-0}=-0.197 (or-0.196) lbs/day)\)
(b) A'(15) = -0.164 (or – 0.163)
The amount of grass clippings in the bin is decreasing at a rate of 0.164 (or 0.163) lbs/day at time t = 15 days.
(c) \(A(t)=\frac{1}{30}\int_{0}^{30}A(t)dt\Rightarrow t=12.415 (or 12.414)\)
(d) L(t) = A(30) + A'(30) • (t-30)
A'(30) = – 0.055976
A(30) = 0.782928
L(t) = 0.5 ⇒ t = 35.054