velocity vector: y=x^3, with x=t, y=t^3 at point t=1

[mattflint50]

Im trying to figure out a Velocity Vector problem on the Calc BC exam, but I have never seen it expressed like this. Can you please help.

The velocity vectore of a particle moving along the curve y=x^3 accordign to x=t and y=t^3 at the point t=1.

Can you please help me.

 

[soroban]

Re: velocity vector

Hello, mattflint50!

The velocity vector of a particle moving along the curve \(\displaystyle y\,=\,x^3\)
according to: \(\displaystyle x\,=\,t\) and \(\displaystyle y\,=\,t^3\) at the point \(\displaystyle t\,=\,1\)

\(\displaystyle \text{We have: }\,\frac{dx}{dt}\,=\,1\) and \(\displaystyle \,\frac{dy}{dt}\,=\,3t^2\)

\(\displaystyle \text{At }t\,=\,1:\;\frac{dx}{dt}\,=\,1,\;\frac{dy}{dt}\,=\,3\)

            B 
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          / :
        v/  : 3
        /   :
       /θ   :
      * - - *
     A   1

\(\displaystyle \text{The velocity vector is: }\,\vec{v}\:=\:\vec{AB} \:= \:\langle r\cdot\cos\theta,\;r\cdot\sin\theta\rangle\)

\(\displaystyle \text{where: }\,r\:=\:|\vec{v}|\:=\:\sqrt{1^2\,+\,3^2}\:=\:\sqrt{10}\)

\(\displaystyle \;\;\text{and: }\,\theta\:=\:\arctan(3)\:\approx\;1.25 \text{ radians.}\)