Integral: find position fcn of particular w/ v(t)=sint-cost

[CalcEqualsUgh]

A particle has velocity v(t) = sint - cost and it's initial position is s(0) = 1. Find the position function of the particle.

So, I know you take the antiderivative of the velocity function to get the position function. And I get s(t)= -cost + sint

And I know I'm supposed to use the fact that s(0)=1 to get the constant.. But I'm not exactly how to do that.

 

[soroban]

Re: Integral

Hello, CalcEqualsUgh!

A particle has velocity: .\(\displaystyle v(t) \:= \:\sin t - \cos t\) and its initial position is: .\(\displaystyle s(0) \,= \,1\)

Find the position function of the particle.

When you integrate, you get "the constant."

. . \(\displaystyle s(t) \;\;=\;\;\int(\sin t - \cos t)\,dt \;\;=\;\;-\cos t - \sin t + C\;\;{\bf[1]}\)

\(\displaystyle \text{We're told that when }t \,= \,0,\:s(0) \,=\,1\)

\(\displaystyle \text{Substitute into }{\bf[1]}: \;-\cos 0 - \sin 0 + C \;=\;1\quad\Rightarrow\quad -1 + C \;=\;1\quad\Rightarrow\quad C \:=\:2\)

\(\displaystyle \text{Therefore: }\;s(t) \;=\;-\cos t - \sin t + 2\)