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$x=t^2-3$

$y=9t$

Assume that a point moves in (x,y) plane according to the above parametric equations. Find for which $t$ the velocity is equal to 9.

I have an idea about this problem, but I'm not sure how to work with it. If we are given the position functions, then the derivative is the velocity. The derivative of both of those with respect to t is $x=2t, y=9$, but I'm not sure what to do after that.

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2 Answers

up vote 4 down vote accepted

$v(t)=(2t,9)$ ( Vector ), then in magnitude get $\|v(t)\| = \sqrt{(2t)^2+9^2}$ ( speed ), thus

$$\|v(t)\| = 9 \Rightarrow \sqrt{(2t)^2+9^2}= 9$$ Now solve for $t$.

EDIT

Magnitude

alt text

$A_{y}$ and $A_{x}$ are the components of vector $A$, the magnitude of vector $A$ is the distance from $ A $ to the origin. By pytagorean theorem the length of $ A $ is given by:

$$A^2=A_{x}^2+A_{y}^2 \Rightarrow A=\sqrt{A_{x}^2+A_{y}^2}$$ ( In magnitude)

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So is pythagorean theorem? How did you know to sum and root them? –  maq Dec 10 '10 at 3:11
    
@f-Prime: Yes, it is the pytagorean theorem. Remember it is the magnitude of a vector. –  Bryan Yocks Dec 10 '10 at 3:18
    
@f-Prime: See the edit. –  Bryan Yocks Dec 10 '10 at 3:49
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Hint: position vector $r=<x(t),y(t)>$. Velocity vector is $dr/dt$ and speed is the magnitude of this vector.

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