Suppose there is a vector $\vec v$ which is a function of time, then will $\dfrac{d}{dt}|\vec v|$ be a vector quantity or a scalar quantity?

Question

Suppose there is a vector $\vec v$ which is a function of time, then will $\dfrac{d}{dt}|\vec v|$ be a vector quantity or a scalar quantity?

I think it should be scalar because, let's assume $\vec v=2t\hat{i}$. Then $|\vec v|=2t$, and $\dfrac{d}{dt}|\vec v| = 2$ which is just a magnitude and has no associated direction.

However, while studying circular motion, I encountered tangential acceleration which is defined as a rate of change of speed. But tangential acceleration has a direction (along the direction of velocity) and thus it is a vector quantity. Thus contradicting what I had said earlier about the derivative of a scalar quantity being a scalar.

I am having trouble figuring out why my reasoning is wrong, please correct me.

Answer 1

The careful mathematics goes like this:

The rate of change of the speed of the particle is given by $$ \frac{dv}{dt} = \frac{d}{dt}\sqrt{\vec{v}\cdot\vec{v}}. $$ Using the chain and product rules of differentiation, we get $$ \frac{dv}{dt} = \frac{1}{2\sqrt{\vec{v}\cdot\vec{v}}}\frac{d}{dt}\vec{v}\cdot\vec{v} = \frac{1}{2v}\left(\frac{d\vec{v}}{dt}\cdot\vec{v}+\vec{v}\cdot\frac{d\vec{v}}{dt}\right) = \frac{1}{2v}\left(\vec{a}\cdot\vec{v}+\vec{v}\cdot\vec{a}\right) = \frac{\vec{a}\cdot\vec{v}}{v} = \vec{a}\cdot\hat{v}, $$ where $\hat{v}$ is the unit vector in the direction of $\vec{v}$, so $\hat{v}$ it is the direction of $\vec{v}$. From this, we can see that since we are dotting the acceleration into the velocity, we get the component of $\vec{a}$ along $\hat{v}$ that leads to changes in speed. This component is what you would call $a_t = \vec{a}\cdot\hat{v}$, and because of the dot-product, it is manifestly not a vector quantity.

Next, we look at how the direction of $\vec{v}$ is changing. Since the direction of $\vec{v}$ is just $\hat{v}$, we want to compute the derivative of $\hat{v}$: $$ \frac{d\hat{v}}{dt} = \frac{d}{dt}\frac{\vec{v}}{v} = \frac{1}{v}\frac{d\vec{v}}{dt} - \vec{v}\frac{1}{v^2}\frac{dv}{dt}, $$ where we again used the product rule (first) and then the chain rule. We rearrange this equation carefully and substitute in for $dv/dt$ from our previous calculation, resulting in $$ \frac{d\hat{v}}{dt} = \frac{1}{v}\left(\vec{a} - (\vec{a}\cdot\hat{v})\hat{v}\right). $$ The quantity in parentheses is exactly the component of $\vec{a}$ perpendicular to the velocity. (You can check orthogonality by taking the dot product of this vector with $\vec{v}$ and finding that it's zero.) The change in direction $d\hat{v}/dt$ therefore depends only on this perpendicular component, which we might call $\vec{a}_r = \vec{a} - (\vec{a}\cdot\hat{v})\hat{v}$.

Answer 2

$a_t=d|v_{t}|/dt$

This only gives the magnitude of the tangential acceleration, overall tangential acceleration is a vector quantity.

Answer 3

$|\vec{v}|$ is the norm of the vector $\vec{v}$, and is a scalar value. If $\vec{v}$ is velocity, $|\vec{v}|$ is speed.

$\frac{\mathrm{d}}{\mathrm{d}t}\vec{v} = \vec{a}$, the acceleration vector.

When we say "tangential acceleration", the direction is "the tangential direction"

$\frac{\mathrm{d}}{\mathrm{d}t} |\vec{v}_t| = |\vec{a}_t|$ is the magnitude of the tangential acceleration.

Answer 4

The time derivative of $ \lvert \vec v \rvert $ is the tangential component of acceleration, which is a scalar quantity, and not the tangential projection, which is a vector. The term ‘tangential acceleration’ is ambiguous and arguably could mean either of these, but they are not the same thing.

Setting physics aside for now, given any vector $ \vec u $ and any nonzero vector $ \vec v $, you can define the component of $ \vec u $ in the direction of $ \vec v $ as a scalar quantity: $ \operatorname { comp } _ { \vec v } \vec u = \vec u \cdot \vec v / \lvert \vec v \rvert $. You can also define the projection of $ \vec u $ along the direction of $ \vec v $ as a vector quantity: $ \operatorname { proj } _ { \vec v } \vec u = ( \vec u \cdot \vec v / \lvert \vec v \rvert ^ 2 ) \, \vec v $. These are related, as $ \operatorname { proj } _ { \vec v } \vec u = ( \operatorname { comp } _ { \vec v } \vec u ) \, \vec v / \lvert \vec v \rvert $, and $ \operatorname { comp } _ { \vec v } \vec u = \pm \lvert \operatorname { proj } _ { \vec v } \vec u \rvert $ (with plus if $ \vec u \cdot \vec v $ is positive, minus if $ \vec u \cdot \vec v $ is negative, and both if $ \vec u \cdot \vec v $ is zero because then both of these quantities are zero).

When $ \vec v $ is a standard basis vector ($ \hat \imath $ or $ \hat \jmath $ in 2 dimensions), then these are the ordinary components; that is, $ \operatorname { comp } _ { \hat \imath } ( a \hat \imath + b \hat \jmath ) = a $, and $ \operatorname { comp } _ { \hat \jmath } ( a \hat \imath + b \hat \jmath ) = b $. In constrast, $ \operatorname { proj } _ { \hat \imath } ( a \hat \imath + b \hat \jmath ) = a \hat \imath $, and $ \operatorname { proj } _ { \hat \jmath } ( a \hat \imath + b \hat \jmath ) = b \hat \jmath $. You can also write $ \vec u = ( \operatorname { comp } _ { \hat \imath } \vec u ) \hat \imath + ( \operatorname { comp } _ { \hat \jmath } \vec u ) \hat \jmath $ and $ \vec u = \operatorname { proj } _ { \hat \imath } \vec u + \operatorname { proj } _ { \hat \jmath } \vec u $. (And this works for any orthonormal basis, not just for the standard basis $ \{ \hat \imath , \hat \jmath \} $.) This is why, even in the general case, we use the word ‘component’. (For the reason why we say ‘projection’, imagine shining a light on $ \vec u $ from a direction perpendicular to $ \vec v $ and observing its shadow on the line through $ \vec v $.)

Now, when $ \vec v $ is the velocity vector of an object in motion, then the direction of $ \vec v $ (assuming that $ \vec v $ is nonzero so that this makes sense) is always tangent to the curve of motion, so $ \operatorname { comp } _ { \vec v } \vec u $ may be called the tangential component of $ \vec u $, and $ \operatorname { proj } _ { \vec v } \vec u $ is the tangential projection of $ \vec u $. If $ \vec u $ is the acceleration $ \mathrm d \vec v / \mathrm d t $ (where $ t $ is time), then by differentiating $ \lvert \vec v \rvert ^ 2 = \vec v \cdot \vec v $, we get $ 2 \lvert \vec v \rvert \, \mathrm d \lvert \vec v \rvert = 2 \vec v \cdot \mathrm d \vec v $, so $ \mathrm d \lvert v \rvert / \mathrm d t = \vec v \cdot \vec u / \lvert \vec v \rvert = \operatorname { comp } _ { \vec v } \vec u $. Thus, the derivative of speed with respect to time is the tangential component of acceleration.

So you're hearing ‘tangential acceleration’ and interpreting this as the tangential projection, which confuses you since that's a vector. But what is really meant (and should be said) is the tangential component of acceleration, and that's a scalar.