A point is moving along the curve $y = x^2$ with unit speed. What is the magnitude of its acceleration at the point $(1/2, 1/4)$?

Question

A point is moving along the curve $y = x^2$ with unit speed. What is the magnitude of its acceleration at the point $(1/2, 1/4)$?

My approach : I use the parametric equation $s(t) = (ct, c^2t^2)$, then $v(t) = s'(t) = (c, 2c^2t)$ and $a(t) = v'(t) = (0, 2c^2)$. Now the point $(1/2, 1/4)$ is reached at time $t = \frac{1}{2c}$, so $v(\frac{1}{2c}) = (c, c)$. Now the unit speed condition gives us $\sqrt{c^2 + c^2} = 1 \implies c = \frac{1}{\sqrt{2}}$. So the magnitude of acceleration is $2c^2 = 1$.

But the answer is $\frac{1}{\sqrt{2}}$. Can someone help me what is wrong in my approach.

Answer 1

What is wrong is that using the parametric coordinates

$$s(t)=(ct,c^2t^2)$$

the point is not moving with unit speed. This can be easily seen by computing the norm of the speed, which is not constant. You need to normalize the speed for all $t$. Then find the time at which the point $(1/4,1/2)$ is reached. And finally compute the acceleration at this point.

Answer 2

You can use Cartesian coordinates for this. You have $$y=x^2\implies \dot{y}=2x\dot{x}$$ Therefore at $x=\frac12$, $\dot{y}=\dot{x}$

For constant speed we have $${\dot{x}}^2+{\dot{y}}^2=1$$ Hence at $x=\frac12$, $\dot{y}=\dot{x}=\frac{1}{\sqrt{2}}$

Differentiating with respect to time the curve again, we have $$\ddot{y}=2{\dot{x}}^2+2x\ddot{x}$$

Therefore at $x=\frac12$, we have $$\ddot{y}=1+\ddot{x}$$

Finally, differentiating the constant speed condition gives $$2\dot{x}\ddot{x}+2\dot{y}\ddot{y}=0\implies \ddot{x}=-\ddot{y}$$

So $\ddot{y}=\frac12$ and $\ddot{x}=-\frac12$ and hence the result.