This problem requires a calculator. Enter answers as decimals.
Grass clippings are placed in a bin, where they decompose. The amount of grass clippings measured in kilograms is modeled by A(t)=7.0(0.95)t, where t is measured in days.
(a) Find the average rate of change of A(t) over the interval 0≤t≤30, in kilograms per day.
(Enter your answer to two decimal places.)
- sin responder
(b) Find the value of A′(15).
(Enter your answer to two decimal places.)
- sin responder
(c) Let L(t) be the linear approximation to A at t=15. Use L(t) to predict the time at which there will be no grass clippings remaining.
(Round to nearest integer.)
- sin responder
(a) To find the average rate of change of A(t) over the interval 0≤t≤30, we need to calculate the difference in A(t) divided by the difference in t.
The formula for average rate of change is:
Average rate of change of A(t) = (A(30) - A(0)) / (30 - 0)
Substituting the values into the formula, we have:
Average rate of change of A(t) = (7.0(0.95)^30 - 7.0(0.95)^0) / (30 - 0)
To solve this equation, we need a calculator.
Calculating A(30):
A(30) = 7.0(0.95)^30 ≈ 1.9859
Calculating A(0):
A(0) = 7.0(0.95)^0 = 7.0(1) = 7.0
Substituting the values back into the average rate of change formula:
Average rate of change of A(t) = (1.9859 - 7.0) / 30
Now, we calculate the result using a calculator:
Average rate of change of A(t) ≈ -0.1843 (rounded to two decimal places)
Therefore, the average rate of change of A(t) over the interval 0≤t≤30 is approximately -0.1843 kilograms per day.
(b) To find the value of A'(15), we need to calculate the derivative of A(t) with respect to t and evaluate it at t = 15.
The equation for A(t) is A(t) = 7.0(0.95)^t.
The derivative of A(t) is obtained by applying the power rule of differentiation:
A'(t) = 7.0(0.95)^t * ln(0.95)
To find A'(15), we substitute t = 15 into the derivative equation and calculate it using a calculator:
A'(15) = 7.0(0.95)^15 * ln(0.95)
Calculating A'(15) using a calculator, we get:
A'(15) ≈ -0.0709 (rounded to two decimal places)
Therefore, the value of A'(15) is approximately -0.0709.
(c) To find the linear approximation L(t) at t = 15, we need to find the equation of the line tangent to the curve of A(t) at t = 15. The equation of a line is given by y = mx + b, where m is the slope and b is the y-intercept.
At t = 15, the slope of the tangent line is A'(15). We have already calculated A'(15) as approximately -0.0709.
We also know that A(15) = 7.0(0.95)^15. Calculating A(15) using a calculator:
A(15) ≈ 1.3327
So, the point (15, 1.3327) lies on the tangent line.
Using the point-slope form of a line, we can find the equation of the tangent line:
y - y₁ = m(x - x₁)
y - 1.3327 = -0.0709(x - 15)
y - 1.3327 = -0.0709x + 1.0635
y = -0.0709x + 2.3962
To find the time at which there will be no grass clippings remaining, we need to find when A(t) = 0.
Setting A(t) = 0 in the equation of the tangent line:
-0.0709x + 2.3962 = 0
Solving for x:
-0.0709x = -2.3962
x ≈ 33.82
Rounding to the nearest integer, the predicted time at which there will be no grass clippings remaining is 34 days.
Therefore, the time at which there will be no grass clippings remaining is estimated to be approximately 34 days.
To solve this problem, we need to find the average rate of change of A(t) over the given interval, the value of A'(15), and use the linear approximation L(t) to predict the time when there will be no grass clippings remaining.
(a) Average rate of change of A(t) over the interval 0≤t≤30:
The average rate of change of a function over an interval is given by the difference in the function values divided by the difference in the input values. In this case, we need to find A(30) - A(0) divided by 30 - 0.
A(30) = 7.0(0.95)^30 ≈ 1.2625 kg
A(0) = 7.0(0.95)^0 = 7.0 kg
Average rate of change = (A(30) - A(0))/(30 - 0) = (1.2625 - 7.0)/30 ≈ -0.2178 kg/day
Therefore, the average rate of change of A(t) over the interval 0≤t≤30 is approximately -0.2178 kg/day.
(b) Value of A'(15):
The derivative of A(t) with respect to t gives us A'(t). To find A'(15), we need to differentiate A(t) with respect to t and evaluate it at t = 15.
A'(t) = 7.0(0.95)^t * ln(0.95)
A'(15) = 7.0(0.95)^15 * ln(0.95) ≈ -0.0149 kg/day
Therefore, the value of A'(15) is approximately -0.0149 kg/day.
(c) Predicting the time when there will be no grass clippings remaining using linear approximation:
To find the linear approximation L(t) to A at t = 15, we need to find the equation of the tangent line to A at t = 15. The equation of a linear approximation is given by L(t) = A(15) + A'(15)(t - 15).
A(15) = 7.0(0.95)^15 ≈ 4.0430 kg
L(t) = 4.0430 - 0.0149(t - 15)
To predict the time when there will be no grass clippings remaining, we need to find the time t when L(t) = 0.
0 = 4.0430 - 0.0149(t - 15)
0 = 4.0430 - 0.0149t + 0.2235
0.0149t = 4.2665
t ≈ 286.21
Rounding to the nearest integer, the predicted time when there will be no grass clippings remaining is approximately 286 days.
Therefore, the answers to the given questions are:
(a) Average rate of change = -0.2178 kg/day
(b) Value of A'(15) = -0.0149 kg/day
(c) Predicted time = 286 days