Need help with limits and derivatives

Question

{(32/5)+2/5cos(πt/4) for 0 ≤ t ≤ 4

W(t) =

{6+(1/8)(t−4)^2 for 4 < t ≤ 9

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The depth of water in a zoo aquarium at a certain point is modeled by the function W defined above, where W(t) is measured in feet and time t is measured in hours.

a) Find W′(7) . Using correct units, explain the meaning of W′(7) in the context of the problem.

b) The graph of W is concave up for 2≤t≤2.5 . Use the line tangent to the graph of W at t=2 to show that W(2.5)≥6 .

c) Find limt→2 (W(t)−t^2−12/5) / t−2 .

Answer 1

a) As typed, the derivative of W for time t=7 can be found by differentiating the second equation:

W'(t) = 1/4 * (t-4) => W'(7) = 1/4(3) = 3/4.

Physically, this means that 7 hours in, the DEPTH (not volume etc.) of the aquarium is changing by 3/4 feet per hour

b) The derivative of W at time t=2 can be found by differentiating the first equation:

W'(t) = -π/10 * sin(πt/4) => W'(2) = -pi/10*1 = -pi/10 (about -.314)

Since W(2) = 32/5 (6.4), and W is concave up, we know the actual value would be more than the value at t=2 plus the change in t multiplied by the numerical value of the derivative of W at t=2. Specifically:

W(2.5) >= W(2) + dW(2)/dt * dt =>

W(2.5) >= 6.4 + ( -π/10 * .5) =>

W(2.5) >= 6.4 - .1571, so W(2.5) >= 6.2ish, which is definitely greater than 6

c) First let's get the expression for the limit sorted out with some algebra:

Simplified, the expression is lim(t->2): (2/5 cos(πt/4) - t^2 + 4) / (t-2).

Trying direct evaluation at t=2 gives us an indeterminate form: 0/0, so we use L'hospital and take the limit of the derivatives of the numerator and denominator:

lim(t->2): - (π/10 sin(πt/4) + 2t) / 1 = -(π/10 + 4) ~ -4.314

It's possible I made some mistakes number crunching, but the derivatives should all be correct.