Can you please check my work?
Two particles start out at t = 0 at he origin. One moves along the positive x-axis at the rate of sin(3pi*t) meters per second and the other moves along the positive y-axis at he rate of 1/(1 + t^2) meters per second. What is the net distance between them after 10 seconds?

Question

Can you please check my work?
Two particles start out at t = 0 at he origin. One moves along the positive x-axis at the rate of sin(3pi*t) meters per second and the other moves along the positive y-axis at he rate of 1/(1 + t^2) meters per second. What is the net distance between them after 10 seconds?

anonymous · Answer

I found the integral of sin(3pi*t) to be (-cos(3pi*t))/3pi

anonymous · Answer

I then evaluated the equation @ t=10 and got -1

anonymous · Answer

Then I did the same for the second equation and got 1.47113

anonymous · Answer

Then I used the distance formula and got 1.7788 as my answer. It this right?

anonymous · Answer

Very close, but not quite! $$ x' = sin(3\pi t) $$$$x=-\frac{cos(3\pi t)}{3\pi} + c$$
Now we apply our initial condition: x(0)=0:
$$x(0)=0=-\frac{1}{3\pi} + c$$$$c=\frac{1}{3\pi}$$Combining this, we see: $$x=\frac{1-cos(3\pi t)}{3\pi}$$ and $$x(10)=\frac{1-cos(30\pi)}{3\pi} = \frac{0}{3\pi}=0$$This may seem odd, but as -1 <= sin(x) <= 1, the speed can be negative implying it's going backwards. Now for y: $$y'=\frac{1}{1+t^2}$$$$y=tan^{-1}(t) + c $$Applying the condition y(0)=0, we see c=0 so $$y=tan^{-1}(t)$$ So $$y(10)=tan^{-1}(10) = 1.471127674$$. As x=0, our value for y(10) is the distance :)

anonymous · Answer

I see what I did wrong. I neglected the constant. Thanks for your help.

anonymous · Answer

Yeah it's an easy mistake to make, you were so close. No problem!