Calculus Problem

Question

Find f(t) if f''(t)=3/sqrt(t) with initial conditions f(4)= -10 and f'(4) = 6 (there is a square root t in the denominator)

f(t) = ___

Answer 1

Give that the second derivative of f(t) --> f"(t) = 3/sqrt(t), we can integrate to achieve the first level of simplification to f'(t)

f'(t) = ∫ f"(t) * dt = ∫3/sqrt(t) * dt = ∫3*t^-1/2 * dt = 3 * t^1/2 / (1/2) + C = 6t^1/2 + C

Using the first derivative initial conditions tells us --> when x = 4 , f' = 6

6 = 6*sqrt(4) + C --> C = -6

f(t) = ∫ f'(t) * dt = ∫(6t^1/2 - 6) *dt = 6t^3/2 / (3/2) - 6t + C = 4t^3/2 - 6t + C

Using the functional value initial condition tells us --> when x = 4, f = -10

-10 = 4*6^3/2 - 6(4) + C --> C = -10 - 24*(sqrt(6) - 1)

f(t) = 4t^3/2 - 6t -10 - 24*(sqrt(6) - 1)