[SOLVED] Differentiation / Integration

• Dec 13th 2009, 07:01 PM
Ife
[SOLVED] Differentiation / Integration
Given the velocity and initial postion of a body moving along a coordinate line at time t, find the body's postion at time t.
$v = cos \frac{ \pi}{2}t, s(0)=1$

I am sure that my answer is wrong, i am thinking that if $s' = cos \frac{ \pi}{2}t$, then $s = sin \frac{ \pi}{2}t+c$, and if s(0)=1, then c=1 so $s = sin \frac{ \pi}{2}t +1$

But i know that if i were to differentiate that same expression, i would get something else not so? (chain rule)... so what should I do here?
• Dec 13th 2009, 07:09 PM
11rdc11
Quote:

Originally Posted by Ife
Given the velocity and initial postion of a body moving along a coordinate line at time t, find the body's postion at time t.
$v = cos \frac{ \pi}{2}t, s(0)=1$

I am sure that my answer is wrong, i am thinking that if $s' = cos \frac{ \pi}{2}t$, then $s = sin \frac{ \pi}{2}t+c$, and if s(0)=1, then c=1 so $s = sin \frac{ \pi}{2}t +1$

But i know that if i were to differentiate that same expression, i would get something else not so? (chain rule)... so what should I do here?

$s = \frac{2\sin{(\frac{\pi t}{2}})}{\pi} + C$
• Dec 13th 2009, 07:10 PM
Ife
Quote:

Originally Posted by 11rdc11

$s = \frac{2\sin{(\frac{\pi t}{2}})}{\pi} + C$

ohhhhh so that i can multiply by the differential of the inner term to get 1... ok.. thanks
• Dec 13th 2009, 07:14 PM
VonNemo19
Quote:

Originally Posted by Ife
Given the velocity and initial postion of a body moving along a coordinate line at time t, find the body's postion at time t.
$v = cos \frac{ \pi}{2}t, s(0)=1$

I am sure that my answer is wrong, i am thinking that if $s' = cos \frac{ \pi}{2}t$, then $s = sin \frac{ \pi}{2}t+c$, and if s(0)=1, then c=1 so $s = sin \frac{ \pi}{2}t +1$

But i know that if i were to differentiate that same expression, i would get something else not so? (chain rule)... so what should I do here?

Hey Ife

$s(t)=\int{v}(t)dt=\int\cos{(\frac{\pi}{2}t)}dt$

Let $u=\frac{\pi}{2}t$, then $\frac{2}{\pi}du=dt$ .

$\Rightarrow{s}(t)=\frac{2}{\pi}\int\cos{u}du=\frac {2}{\pi}\sin{u}+C=\frac{2}{\pi}\sin(\frac{\pi}{2}t )+C$

Now

$s(0)=1\Rightarrow1=\frac{2}{\pi}\sin(\frac{\pi}{2} \cdot0)+C\Rightarrow{C}=1$