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An object moves along a path given by the equation $$ y(x)=2x^{2}-3x-11$$ with a constant speed of 5m/s. Find the velocity at x=2.

My approach: We know that the speed of the object is given by $$ 25=v^{2} = v_{x}^2+v_y^2$$ Therefore $$25 =v_x^2(1+(\frac{v_x}{v_y})^2)=(v_x^2(1+(\frac{dy/dt}{dx/dt})^2)=(v_x^2(1+(\frac{dy}{dx})^2)$$ now, we know that $$ \frac{dy}{dx}=4x-3$$ so $$v_x^{2}=\frac{25}{(1+(4x-3)^2)}$$ and $$v_x=\sqrt\frac{25}{26}$$ now $$v_y=\sqrt{25-\frac{25}{26}}$$ and finally $$\vec v (2)=\frac{5}{\sqrt{26}}(\vec i + 5\vec j)$$ is my approach correct? Also, let's take $$x=f(t)$$ is there any way to parametrize the curve with f so that we can take $$\vec v = df/ dt $$ as the velocity?

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I think it is much simpler than what you are doing. As you mention, $\frac{dy}{dx} = 4x-3$. So we know that the tangent vector to the curve (no matter how it is parameterized) points in the direction of the vector $(1,4x-3)$, since this will be tangent to the curve. When $x=2$, this is $(1,5)$. So the velocity vector points in this direction. We just have to re-scale it so the length is $5$. The length of $(1,5)$ is $\sqrt{26}$, so the velocity should be $\frac{5}{\sqrt{26}}(1,5)$.

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  • $\begingroup$ If so, why are our answers different? 5/√ 26!=5/6? Which part did I get wrong? $\endgroup$ – Mario Aldean May 20 at 0:41
  • $\begingroup$ When you solved for $v_x^2$, you have $(1+(4x-3))^2$ in the denominator, instead of $1+(4x-3)^2$. $\endgroup$ – Nick May 20 at 0:44
  • $\begingroup$ Thank you, typo ended up trashing the question entirely. $\endgroup$ – Mario Aldean May 20 at 0:55

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