# Using centripetal acceleration to find back the velocity magnitude at $t+dt$

Considering a circular motion with no angular acceleration. How can you find the same magnitude for the velocity vector at different time using the formula $$v_{t} = v_0 + a.t$$ with vectors?

The acceleration vector $$\vec{a} = \vec{a_c} + \vec{a_t}$$ where $$\vec{a_c}$$ is the centripetal acceleration $$a_c$$ = $$\dfrac{-v^2}{r}$$ and $$\vec{a_t}$$ is the tangential acceleration if the tangential speed change.

On every diagram I've seen so far, the radial acceleration vector $$\vec{a_r}$$ and the tangential velocity $$\vec{v_t}$$ are perpendicular and their vector tails share the same point, then how could $$\vec{v_{(t_0+dt)}}$$ be the same magnitude than $$\vec{v_{(t_0)}}$$ ? How can I prove a hypothenuse be the same length as one of its component? In my book they speak of radial acceleration instead of centripetal but if I'm right, I tried this : $$\vec{v_{(t_0+dt)}} = v_{r_{(t_0)}}\vec{r} + v_{t_{(t_0)}}\vec{\theta} + (a_r.dt)\vec{r} + (a_t.dt)\vec{\theta}$$

$$\vec{r}$$ is the unit vector in the radius vector's direction which goes to the point p at $$t_0$$

and $$\vec{\theta}$$ is the unit vector of the tangent at the point p and in the positive direction counterclockwise and perpendicular to $$\vec{r}$$

the scalar $$v_r$$ is zero and with no angular acceleration $$a_t$$ is zero. Is this equation correct? $$\mid\mid\vec{v_{(t_0+dt)}}\mid\mid = (v_{t_{(t_0)}}^2 + (a_r.dt)^2)^{\frac{1}{2}}$$ I've just seen the derivation leading to $$a_c$$ = $$\dfrac{-v^2}{r}$$ but in the equation above any magnitude for $$a_r$$ would give $$v_{(t_0+dt)} = v_{t_{(t_0)}}$$ since $$dt$$ is small enough.

It's very easy to do: \begin{align} |\mathbf{v}(t+dt)|^2&= \left|\mathbf{v}(t)+\frac{d\mathbf{v}}{dt}dt\right|^2\\ &=\left|v(t)\boldsymbol{\hat\theta}+\mathbf{a}(t)dt\right|^2\\ &=\left|v(t)\boldsymbol{\hat\theta}+a(t)dt\mathbf{\hat r}\right|^2\\ &=v^2(t)+2v(t)a(t)dt \boldsymbol{\hat\theta}\cdot\mathbf{\hat r}+a^2(t)(dt)^2 \end{align} Since $$\boldsymbol{\hat\theta}\cdot\mathbf{\hat r}=0$$ and $$(dt)^2$$ is negligible, $$|\mathbf{v}(t+dt)|^2=v^2(t)$$ It works for any acceleration as long as it's perpendicular to velocity.
There is a general argument that is beautifully simple. The (squared) speed is given by $$\vec{v}\cdot\vec{v}$$. Now, let's consider the rate of change of (squared) speed:
\begin{align} \frac{d}{dt} (\vec{v}\cdot\vec{v})&= \frac{d\vec{v}}{dt}\cdot\vec{v}+ \vec{v}\cdot\frac{d\vec{v}}{dt} \\ &= 2 \frac{d\vec{v}}{dt}\cdot\vec{v}\\ &= 2 \vec{a}\cdot\vec{v} \end{align}