Calculus limits

Question

W(t)={17/2−3/2cos(πt/6) for0≤t≤6

10−1/5(t−6)^2. for6<t≤10

The depth of a river 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.

c) Find  limt→2

( W(t)−t^3+1/4) / (t−2)

Answer 1

First, we see that W(2) = 31/4. Plugging into our expression, we have

(W(t) - t³ + 1/4) / (t - 2) = (31/4 - 8 + 1/4) / 0 = 0 / 0

Therefore, we need to use L'Hopital's Rule. Taking the derivative of the numerator and denominator, we have:

(W'(t) - 3t²) / 1 = (3/2)(π/6)sin(πt/6) - 3t²

Therefore, the limit is

(3/2)(π/6)sin(π/3) - 12 = (√3π/8) - 12

Answer 2

W(t)={17/2−3/2cos(πt/6) for0≤t≤6

Find  limit→2

( W(t)−t^3+1/4) / (t−2)

(17/2−3/2cos(πt/6)−t^3+1/4) / (t−2) as t approaches 2.

and W(t) approaches -5 feet