# Confused about the answer of AP BC 2010 # 3a

A particle is moving along a curve so that its position at time $t$ is $(x(t),y(t)),$ where $x(t) = t^2-4t+8$ and $y(t)$ is not explicity given. Both $x$ and $y$ are measured in meters, and $t$ is measured in seconds. It is known that $\frac{\mathrm{d}y}{\mathrm{d}t} = te^{t-3} - 1$.

(a) Find the speed of the particle at time $t = 3$ seconds.

Answer given: Speed = $\sqrt{\left(x^{\prime}\left(3\right)\right)^2 + \left(y^{\prime}\left(3\right)\right)^2} = 2.828$ meters per second.

Why is the answer not $$\frac{\mathrm{d}y}{\mathrm{d}x}\Bigg|_{x=3}$$

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The formatting in your answer messed up a $. – Amzoti Dec 26 '12 at 1:12 ## 3 Answers Definition: Let the position vector for a particle in the xy-plane be$r(t) = xi + yj = f(t)i + g(t)j,$where$t$is the time, and the scalar functions$f$and$g$have first and second derivatives. The velocity, speed and acceleration of the particle at time$t$are as follows: Velocity:$v(t) = r'(t)$=$\frac{dx}{dt}i$+$\frac{dy}{dt}j$Speed:$s(t) = \|v(t)\|$=$\|r'(t)\|$=$\sqrt{\mathstrut\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^{2}}$Acceleration:$a(t)$=$v'(t)$=$r''(t)$=$\frac{dx^{2}}{dt}i$+$\frac{dy^{2}}{dt}j$In your problem, you are already given$\frac{dy}{dt}$, so you need to compute$\frac{dx}{dt}$and use the formula for speed that is given above at time$t = 3$. You will get the answer you cited in the problem. Clear? Regards - +1 Hello!! TGIF!!$\;\;$:$\tiny\nabla$)$\;\;$– amWhy May 11 '13 at 0:15 @amWhy: Indeed! Got home about an hour ago and it is good being home! This reminded me how much I miss the University environment as a place of exploration, learning and growth. I feel warped, but just answered a question and another one is in my purview. Hope you are having a brilliant day, as I currently feel warped from 350 miles of driving! Regards – Amzoti May 11 '13 at 0:17 Note: You do not need$y(t)$. $$x' = \dfrac{dx}{dt}\quad \text{and} \quad y' = \dfrac{dy}{dt}.$$ You are given$y'$; you need only compute$x'$and use the formula for speed: $$\text{speed}\;= \sqrt{(x')^{2} + (y')^{2}}.\tag{1}$$ Then evaluate at time$t = 3$.$(1)\;$Recall$\dfrac{dx}{dt}$and$\dfrac{dy}{dt}$represent, respectively, the rate of change of the$x$-coordinate and$y$-coordinate with respect to time$t$. The slope of the curve in the$x$-$y$plane is$\dfrac{dy}{dx}$, and this can be computed as$\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$, when$dx/dt\ne 0$. But recall that the speed of a particle along the parametric curve$(x(t),y(t))$is defined by$\dfrac{ds}{dt},\;$where$\;s= \textrm{arc length}$. This is how we can get that$\text{speed} =\dfrac{ds}{dt} = \sqrt{[x′(t)]2+[y′(t)]} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$. - Because speed at time$t=3$it given by $$\sqrt{[x'(t)]^2+[y'(t)]^2}\Bigg\vert_{t=3}=\sqrt{(2t-4)^2+(te^{t-3}-1)^2}\Bigg\vert_{t=3}=2\sqrt{2}\approx 2.82843.$$ Since speed is inherently a time derivative, why would you take${dy\over dx}$? This would be used if you were interested, for example, in the slope of the parametric curve$(x(t),y(t))$in the$x$-$y$plane. - Just to make sure I understand. dy/dx would give you the velocity at a point, but in this reguard, we have the factor of time, so we must use the head-tail method to find the lenght of the derivative vectors. – yiyi Dec 26 '12 at 1:25 Just remember$dx/dt$and$dy/dt$represent the rate of change of the$x$-coordinate and$y$-coordinate with respect to$t$. Then the slope of the curve in the$x$-$y$plane is$dy\over dx$, but we can get at this via${dy\over dx}={dy/dt\over dx/dt}$, for$dx/dt\not=0$. Finally, I'm not sure in what sense you mean velocity here. Recall that the speed of a particle along the parametric curve$(x(t),y(t))$was defined as${ds\over dt}$where$s$was arc length. This is where the proof that speed equals$\sqrt{[x'(t)]^2+[y'(t)]^2}\$ came from. – JohnD Dec 26 '12 at 1:41
Yes, it was. Just forgot which "Space" i was in. parametric not the normal x y plane – yiyi Jan 2 '13 at 12:56