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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
up vote 8 down vote accepted


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.



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+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

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.

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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

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 $\text{speed} =\dfrac{ds}{dt} = \sqrt{[x′(t)]2+[y′(t)]} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$.

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