Cal word prob??

Question

The weight in kilograms of rocket fuel in a rocket launcher is given by W = (1/t) - (4/t^2), where t is time in seconds. At time t = 10 seconds, the amount of fuel in the launcher is decreasing at a rate of _____ kilograms per second

Answer 1

I'm pretty sure you have to find the value of f'(10).

So the original equation is

w = 1/t - 4/t^2

The derivative = -t^(-2) - (-8t^-3)

Substitute 10 in...

-(10)^(-2) - (-8(10)^(-3)) = -.01 - (-.008) = -.002 kg/sec

Answer 2

dW/dt = -1/t^2 - (-8t/t^4) = -1/t^2 + 8/t^3
Whenn t = 1, dW/dt = -1/10^2 +8/10^3 = -.002 kg/sec