Application of derivatives problem?

[mirs]

Homework Statement

A biologist determines experimentally that the number of calories burned by a salmon swimming a distance d in miles upstream against a current v0 in miles per hour is given by

Energy = kdv^5/v − v0

where v is the salmon’s swimming speed relative to the water it is in. This means that the salmon’s progress upstream is at the rate of v − v0 miles per hour, so that the distance d is covered in a time of

t=d/v − v0

If v0 = 2 mph and d = 20 miles, and the salmon, being smart, swims so as to minimize the calories burned, how many hours will it take to complete the journey?

Homework Equations

Requires derivatives

The Attempt at a Solution

I really just don't understand where to start. I subbed t into the equation and ended up with ktv^5/v-v0, took the derivative of that with respect to t and got 5kv^4 but I have no idea what I really don't know where to go from there.

 

[LCKurtz]

Homework Statement

A biologist determines experimentally that the number of calories burned by a salmon swimming a distance d in miles upstream against a current v0 in miles per hour is given by

Energy = kdv^5/v − v0

Do you mean $\frac{kdv^5} v - v_0$, which is what you wrote, or $\frac{kdv^5}{v − v0}$?

where v is the salmon’s swimming speed relative to the water it is in. This means that the salmon’s progress upstream is at the rate of v − v0 miles per hour, so that the distance d is covered in a time of

t=d/v − v0

Same problem here. Obviously you mean $\frac d {v-v_0}$. Use parentheses!

If v0 = 2 mph and d = 20 miles, and the salmon, being smart, swims so as to minimize the calories burned, how many hours will it take to complete the journey?

Homework Equations

Requires derivatives

The Attempt at a Solution

I really just don't understand where to start. I subbed t into the equation and ended up with ktv^5/v-v0, took the derivative of that with respect to t and got 5kv^4 but I have no idea what I really don't know where to go from there.

Try finding the critial value of $v$ using the energy equation. Then use that to get the time of travel.

 

[mirs]

Yes! I meant (kdv^5)/(v−v0).

 

[mirs]

Try finding the critial value of $v$ using the energy equation. Then use that to get the time of travel.

OK, so I'm guessing I don't plug t into the energy equation in the first place? because I did that and found the derivative to be 5ktv^4

 

[LCKurtz]

Try finding the critial value of $v$ using the energy equation. Then use that to get the time of travel.

OK, so I'm guessing I don't plug t into the energy equation in the first place? because I did that and found the derivative to be 5ktv^4

Why are you guessing? Why not just try what I suggested?