(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.