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.