asked by @fabiolar1 •
almost 2 years ago •
Physics
• 5 pts

*v(t)=3t^2+2e^(2t)+5cos(4t)+2 where t1=2 t2=5*

First, you need to find the derivative of the function, THEN you want to evaluate it at the specific times. Notice that if you were to just plug the numbers in first, your velocity equation would be a single number, and the derivative of that number would always be 0, so that can't be the way to do the problem. You need to take the derivative while it is still a function, BEFORE plugging in specific time values. So let's do that.

The first term is easy to take the derivative of since it is a polynomial:

d/dt (3t^2) = (2*3)t = 6t

The second term is more complicated because you need to use the chain rule on it. If it were simply e^t, the derivative would be e^t. But since it's e^(2t), you need to do chain rule on the exponent. So

d/dt (2e^(2t)) = 2e^(2t)*(2) = 4e^(2t)

For the third term, once again we need to use chain rule. The normal derivative of cos(t) is -sin(t), so our derivative is

d/dt (5cos(4t)) = -5sin(4t)*(4) = - 20sin(4t)

Finally, the last term is a constnat, so that just becomes 0 when we take the derivative. So, our full derivative function is just the sum of all the individual derivatives we did:

dv/dt = 6t + 4e^(2t) - 20sin(4t)

Evaluated at t1 = 2, we have

dv/dt (2) = 6(2) + 4e^(2*2) - 20sin(4*2) = 210.6

(NOTE: Don't forget to make sure you calculator is in radians -- the input of sine is not in degrees)

Evaluated at t2 = 5, we have

dv/dt (5) = 6(5) + 4e^(2*5) - 20sin(4*5) = 88,118

answered by @doug •
almost 2 years ago

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