# Finding the magnitude of displacement with a given parametric velocity function

I am struggling to get the correct answer for the question:

What is the magnitude of the displacement of a particle moving in the $$x$$-$$y$$ plane with the velocity vector given by $$v(t) = (e^{\sin t}, 5t^2)$$ for time $$t \ge 0$$ between time $$t=1$$ and $$t=2$$.

I attempted to take the integral of the speed function $$\int_1^{2} (\sqrt{e^{2\sin t}+25t^4}) \, \mathrm{d}t \\$$ since the velocity of the particle is always positive on the interval and got the answer $$11.992$$. However, apparently the correct answer is $$11.954$$, but I can't find a way to come about that answer.

• What you need is $\left\| \int_1^2 v \mathrm{d}t\right\|$, not $\int_1^2\left\| v \right\|\mathrm{d}t$. Mar 16, 2019 at 23:57
• @coreyman317 how would I go about evaluating that integral since $v(t)$ is a parametric function? Mar 17, 2019 at 4:11
• @ChargeShivers how would I go about evaluating that integral since v(t) is a parametric function? Mar 17, 2019 at 4:27
• Let $v = (v_x, v_y)$, then $\int v \mathrm{d}t = \left( \int v_x \mathrm{d}t, \int v_y\mathrm{d}t\right)$, and so $\left\| \int v \mathrm{d}t \right\| = \sqrt{ \left( \int v_x \mathrm{d}t \right)^2 + \left( \int v_y \mathrm{d}t \right)^2 }$. Mar 18, 2019 at 3:39