# Thread: First and second derivative of vector function?

1. ## First and second derivative of vector function?

$\displaystyle f (t) = e^t cos(t)i-sin(2t)j + 3k.$

It has been a while since I have taken derivatives, but I know the basic rules... I am just having trouble understanding what is constant and what is a variable. Could you help me out with the second derivative of this function?

2. Originally Posted by Mattpd
$\displaystyle f (t) = e^t cos(ti)-sin(2tj) + 3k.$
First, this should be $\displaystyle e^t cos(t)i- sin(2t)j+ 3k$ because cosine and sine of a vector is not defined!
It has been a while since I have taken derivatives, but I know the basic rules... I am just having trouble understanding what is constant and what is a variable. Could you help me out with the second derivative of this function?
The derivative, with respect to t, of f(t)i+ g(t)j+ h(t)k, is f'(t)i+ g'(t)j+ h'(t)k.

3. So pull out the i, j, and k as constants and take the derivative of the rest, but how do I treat the "t"?

4. Originally Posted by Mattpd
So pull out the i, j, and k as constants and take the derivative of the rest, but how do I treat the "t"?
$\displaystyle f(t) = e^t \cos t$, $\displaystyle g(t) = - \sin (2t)$ and $\displaystyle h(t) = 3$.

Are you saying that you don't know how to differentiate these functions?

5. f(t)=-e^t sin(t) + e^tcos(t)
g(t)=-2cos(2t)
h(t)=0
???

6. Originally Posted by Mattpd
f'(t)=-e^t sin(t) + e^tcos(t)
g'(t)=-2cos(2t)
h'(t)=0
???
I assume that you meant to include the red dashes to denote derivative ...? Your derivatives are correct. Now you have to differentiate again to get the second derivative of the vector function.

By the way, note that the derivatives by themselves are not the answer to the question.

7. Thanks, I got it from here. The question is for a grade so I just wanted to make doubly sure that I am correct.