Hi there!
a)
We can approximate g'(3) by finding the slope of g(t) over an interval containing t = 3.
We can use the endpoints t = 1 and t = 5 min for the best estimate.
[tex]slope = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20.5-15.1}{5-1} = \boxed{1.35 \frac{ft^3}{min^2}}[/tex]
This means that the rate of grain being added to the silo is INCREASING at a rate of 1.35 ft³/min². (Or in other words, the grain is being poured at an increasingly greater rate)
b)
The total amount of grain added is the integral of g(t), so:
[tex]\text{total amount of grain} = \int\limits^8_0 {g(t)} \, dt[/tex]
We can do a right riemann sum by using the right endpoints (t = 1, t = 5, t = 6, t = 8) to calculate.
Riemann sums are essentially rectangles added up to calculate an approximate value for the area under a curve.
The bases are the spaces between each value in the chart, while the heights are the values of g(t).
Using the intervals and values in the chart:
[tex]1(15.1) + 4(20.5) + 1(18.3) + 2(22.7) = \boxed{160.8 ft^3}[/tex]
c)
We can subtract the two integrals to find the total amount of unspoiled grain.
With g(t) being fresh grain and w(t) being spoiled grain, let y(t) represent unspoiled grain.
[tex]y(t) = \int\limits^8_0 {g(t)} \, dt - \int\limits^8_0 {w(t)} \, dt[/tex]
Use a calculator to evaluate:
[tex]y(t) = 160.8 - \int\limits^8_0 {w(t)} \, dt = 160.8 - 99.05 = \boxed{61.749 ft^3}[/tex]
d)
We can do the first derivative test to determine whether the amount of grain is increasing or decreasing. (Whether the first derivative is positive or negative at this value).
For the above integral, we know that the derivative is:
[tex]y'(t) = g(t) - w(t)[/tex]
Plug in the values for t = 6:
[tex]w(6) = 32 \cdot \sqrt{sin(\frac{\pi (6)}{74})} = 16.06[/tex]
[tex]y'(6) = g(6) - w(6) = 18.3 - 16.06 = 2.23 \frac{ft^3}{min}[/tex]
This value is POSITIVE, so the amount of unspoiled grain is INCREASING.
