# Prove that a non-zero acceleration is perpendicular to a constant speed

Take the differentiable vector function $\vec{v}(t)$ (a velocity vector). If its speed, $|\vec{v}(t)|=constant$, then prove that at any point which $\frac{d\vec{v}}{dt}$ is non-zero, $\frac{d\vec{v}}{dt}$ is perpendicular to $\vec{v}(t)$.

In other words, if a velocity vector has constant speed, show that whenever its acceleration vector is non-zero, it is perpendicular to that velocity vector.

Intuitively this seems clear, since whenever the speed is remains constant, the acceleration is 0. Why though are the velocity and acceleration vectors perpendicular though? Because they share no intersection when the acceleration is non-zero since the speed is constant?

I'm assuming we should somehow show that $\vec{v} \cdot \vec{a} = 0$. The thing is, I can't figure out what knowing that the speed is constant tells us about the velocity. It could still be anything (like some trig functions squaring up and becoming 1 through an identity)?

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The absolute value bars should be only around $v(t)$ in the first line; and it should be "its" (the possessive), not "it's" (which means 'it is'). – Arturo Magidin Feb 1 '11 at 5:06
Thanks, I fixed that. – fdart17 Feb 1 '11 at 5:07
It is not always true that if the speed is constant, the acceleration is 0. The speed is the magnitude of the velocity. For example, if the path is a circle at uniform speed, the acceleration is not zero, but $\frac{v^2}{r}$ directed inward. The change in speed is the acceleration projected onto the velocity axis. This is the intuitive content of Arturo Magidin's answer. – Ross Millikan Feb 1 '11 at 5:18
You should clearly distinguish between the speed $v$, a scalar which is measured by means of a speedometer, and the velocity vector ${\bf v}$. The connection between the two is the following: $v=|{\bf v}|$. – Christian Blatter Feb 3 '11 at 8:56
Why did you start the bounty while you already have three answers, each of them proves the question in different way. Generally, all of them rely on the following facts: (a) acceleration is a limit of an infinitesimal change of velocity, (b) since velocity magnitude is constant, all velocity vectors lie an a circle, giving that limit direction of infinitesimal change of velocity lies along the circle tangent, thus being perpendicular to radius (that is, to velocity). Since the velocity vector is tangent to trajectory, thus acceleration is perpendicular to trajectory. – mbaitoff Feb 3 '11 at 9:51

You can apply the product rule when differentiating $\vec v\cdot\vec v =$constant.

The intuitive idea is that $\vec v(t)$ traces out a curve on a sphere centered at the origin (i.e., picture $\vec v(t)$ as a moving radius vector), while $\vec v'(t)=\vec a(t)$ is tangent to the sphere, and hence perpendicular to the radius at the point of tangency, namely $\vec v(t)$.

There is also a geometric description of what this says in terms of the original curve, say $\vec x(t)$, of which $\vec v(t)$ gives the velocity. Since $\vec v(t)$ gives a vector tangent to the trajectory of $\vec x(t)$, and in your case $\vec a(t)$ is perpendicular to $\vec v(t)$, $\vec a(t)$ is perpendicular to the trajectory of the original curve $\vec x(t)$.

Acceleration can occur for 2 reasons. In general, the component of $\vec a(t)$ in the direction of the trajectory (or opposite this direction) tells you how the speed is changing, while the component of $\vec a(t)$ perpendicular to the trajectory tells you how the direction is changing. So, again, in case the speed is constant, $\vec a(t)$ is perpendicular to the curve. Its direction tells you which direction $\vec x(t)$ is turning, while its magnitude tells you how quickly $\vec x(t)$ is changing direction. This last quantity is a constant multiple of the curvature of $\vec x(t)$.

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Ross gives a situation in which there is another intuitive interpretation, uniform circular motion. The interpretation I give basically asks you to forget that $v$ is the velocity of some other curve, and to think of it as based at the origin and tracing out a curve on a sphere. The point is that any curve with constant length is perpendicular to its derivative. – Jonas Meyer Feb 1 '11 at 5:35
And Jonas has it right as well. If $v\cdot v=constant$ and you take a time derivative, you get $2v \cdot a=0$ – Ross Millikan Feb 3 '11 at 5:30
Ah I see. I will probably accept this later. – fdart17 Feb 3 '11 at 13:32

The shortest way is to notice that:

$$v^2=\vec{v}\cdot\vec{v}$$

After differentiating: $$2v\frac{dv}{dt}=2\vec{v}\cdot\frac{d\vec{v}}{dt}$$ Done.

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Why are we done? If $|\vec{v}(t)|$ is the speed, what's $\frac{d}{dt}|\vec{v}(t)|$? Because the acceleration vector is $\frac{d\vec{v}}{dt}$ I edited my question with the correct vector notation. – fdart17 Feb 3 '11 at 13:22
This is a correct and elegant answer. Notice that the plain v in the left side is a scalar, the magnitude of the (vectorial) velocity. Hence the left side is cero (the magnitude of the velocity is a constant, hence it's derivative is zero). Hence, the right side (scalar product of two vectors) must be zero, hence v a = 0 – leonbloy Feb 3 '11 at 16:46

If $v(t) = \bigl(x_1(t),x_2(t),\ldots,x_n(t)\bigr)$ (I don't know what dimension you are working on), then the speed is given by $$||v(t)|| = \sqrt{ x_1(t)^2 + \cdots + x_n(t)^2}.$$ Since this is constant, say equal to $k$, then you have $$k^2 = x_1(t)^2 + \cdots + x_n(t)^2.$$

Now take derivatives with respect to $t$. Compare that with $v\cdot a$ (remembering that $a(t) = v'(t)$).

(Of course, if $\frac{dv}{dt}$ is zero, then it is also perpendicular to $v(t)$, since $\mathbf{0}$ is perpendicular to every vector...)

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Lemma 1: Velocity vector is tangent to the velocity curve.

Then consider the acceleration as a limit of velocity difference. Since the magnitude of the velocity vector is constant, the velocity vector traverse the circle of a constant radius. The difference of the two infinitely close velocity vectors is perpendicular to the radius, hence is perpendicular to velocity vector. Here the solution follows immediately.

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