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Differentiating the Squared Magnitude of a Vector Function

Use the dot-product rule to differentiate a vector function’s squared magnitude and interpret the result in practical calculations.

Beginner 8 min read

What You'll Learn

  • Express the squared magnitude of a vector function as its dot product with itself.
  • Derive the formula for its derivative from the dot-product rule.
  • Calculate squared-magnitude derivatives using both dot products and components.
  • Check derivative calculations by expanding the squared magnitude directly.
  • Use the formula to identify when a changing vector has constant magnitude.
  • Avoid confusing the derivative of squared magnitude with the derivative of magnitude.

Why Differentiate the Squared Magnitude?

A vector function can change in direction, length, or both. Its squared magnitude is a scalar function that tracks one part of this change: for a vector \(\mathbf{r}(t)\), the quantity \(\lVert\mathbf{r}(t)\rVert^2\) measures the square of its length. Differentiating this scalar gives a compact way to describe how the vector’s length is changing, without first differentiating a square root.

The key is to view a squared magnitude as a dot product of a vector with itself. The dot-product rule then gives the derivative directly. This also reveals a useful geometric fact: if the magnitude stays constant, the vector and its derivative are perpendicular.

Definition: For a vector \(\mathbf{r}(t)\) in Euclidean space, its magnitude is \(\lVert\mathbf{r}(t)\rVert=\sqrt{\mathbf{r}(t)\cdot\mathbf{r}(t)}\). Its squared magnitude is \(\lVert\mathbf{r}(t)\rVert^2=\mathbf{r}(t)\cdot\mathbf{r}(t)\).

Here \(\cdot\) denotes the Euclidean dot product. For example, if \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\), then \(\lVert\mathbf{r}(t)\rVert^2=x(t)^2+y(t)^2+z(t)^2\). We will assume that \(\mathbf{r}\) is differentiable on an open interval of parameter values, so its derivative \(\mathbf{r}'(t)\) exists there.

The Squared-Magnitude Derivative Rule

The dot-product rule states that if \(\mathbf{u}(t)\) and \(\mathbf{v}(t)\) are differentiable vector functions, then

$$ \frac{d}{dt}\big(\mathbf{u}(t)\cdot\mathbf{v}(t)\big) = \mathbf{u}'(t)\cdot\mathbf{v}(t) + \mathbf{u}(t)\cdot\mathbf{v}'(t). $$

Apply this rule with both functions equal to \(\mathbf{r}(t)\). The two resulting dot products are equal because the Euclidean dot product is symmetric.

Theorem: If \(\mathbf{r}(t)\) is a differentiable vector function, then $$ \frac{d}{dt}\lVert\mathbf{r}(t)\rVert^2 = 2\mathbf{r}(t)\cdot\mathbf{r}'(t). $$

Why the Formula Works

Since the squared magnitude is the dot product of the vector with itself, the dot-product rule applies directly. In particular, neither copy of \(\mathbf{r}\) is treated as a constant: each contributes a derivative term.

$$ \begin{aligned} \frac{d}{dt}\lVert\mathbf{r}(t)\rVert^2 &=\frac{d}{dt}\big(\mathbf{r}(t)\cdot\mathbf{r}(t)\big)\\ &=\mathbf{r}'(t)\cdot\mathbf{r}(t) +\mathbf{r}(t)\cdot\mathbf{r}'(t)\\ &=2\mathbf{r}(t)\cdot\mathbf{r}'(t). \end{aligned} $$

The final equality follows from symmetry: \(\mathbf{r}'(t)\cdot\mathbf{r}(t)=\mathbf{r}(t)\cdot\mathbf{r}'(t)\). This argument works in any finite-dimensional Euclidean space and does not depend on the vector having exactly three components.

Worked Example: A Polynomial Vector Function

Let \(\mathbf{r}(t)=\langle t^2,3t,1-t\rangle\). Find the derivative of its squared magnitude, then check the result by expanding the squared magnitude.

First differentiate each component:

$$ \mathbf{r}'(t)=\langle 2t,3,-1\rangle. $$

Use the dot-product formula. The dot product is

$$ \begin{aligned} \mathbf{r}(t)\cdot\mathbf{r}'(t) &=(t^2)(2t)+(3t)(3)+(1-t)(-1)\\ &=2t^3+9t-1+t\\ &=2t^3+10t-1. \end{aligned} $$

Therefore,

$$ \frac{d}{dt}\lVert\mathbf{r}(t)\rVert^2 =2(2t^3+10t-1)=4t^3+20t-2. $$

Check by calculating the squared magnitude first:

$$ \begin{aligned} \lVert\mathbf{r}(t)\rVert^2 &=(t^2)^2+(3t)^2+(1-t)^2\\ &=t^4+9t^2+1-2t+t^2\\ &=t^4+10t^2-2t+1. \end{aligned} $$

Differentiating this scalar expression gives \(4t^3+20t-2\), which agrees with the dot-product calculation.

Reading the Result

The formula measures the rate of change of the squared length. At a given parameter value, the sign of \(\mathbf{r}(t)\cdot\mathbf{r}'(t)\) tells whether the squared magnitude is increasing or decreasing there: a positive dot product gives a positive derivative, a negative dot product gives a negative derivative, and a zero dot product gives an instantaneous rate of zero.

A particularly useful case occurs when the magnitude is constant on an interval. Then its square is also constant, so its derivative is zero throughout that interval. The theorem consequently gives \(\mathbf{r}(t)\cdot\mathbf{r}'(t)=0\). Thus the vector is perpendicular to its derivative at each such parameter value. This conclusion requires constant magnitude on an interval, not merely that the magnitude happens to have the same value at two separate parameter values.

Worked Example: A Vector of Constant Magnitude

Let \(\mathbf{r}(t)=\langle 3\cos t,3\sin t\rangle\). Use the squared-magnitude formula to show that the magnitude is constant and that \(\mathbf{r}(t)\cdot\mathbf{r}'(t)=0\).

First compute the derivative:

$$ \mathbf{r}'(t)=\langle -3\sin t,3\cos t\rangle. $$

Now calculate the dot product of the function with its derivative:

$$ \begin{aligned} \mathbf{r}(t)\cdot\mathbf{r}'(t) &=(3\cos t)(-3\sin t)+(3\sin t)(3\cos t)\\ &=-9\cos t\sin t+9\sin t\cos t\\ &=0. \end{aligned} $$

Therefore the derivative of the squared magnitude is

$$ \frac{d}{dt}\lVert\mathbf{r}(t)\rVert^2 =2\mathbf{r}(t)\cdot\mathbf{r}'(t)=2(0)=0. $$

Check directly: \(\lVert\mathbf{r}(t)\rVert^2=9\cos^2t+9\sin^2t=9(\cos^2t+\sin^2t)=9\). Its derivative is \(0\), as required. The magnitude is \(\sqrt{9}=3\) for every \(t\).

Computing at a Specific Parameter

The identity can be used either to find a derivative as a function of the parameter or to evaluate it at a particular parameter value. For a specific value, substitute into both the vector and its derivative, form the dot product, and multiply by two. Keeping the steps in this order makes it easier to check signs and arithmetic.

Worked Example: Evaluating the Rate at \(t=0\)

Let \(\mathbf{r}(t)=\langle e^t,e^{-t},t\rangle\). Find the derivative of the squared magnitude at \(t=0\), and check it by differentiating the expanded expression.

Differentiate the components:

$$ \mathbf{r}'(t)=\langle e^t,-e^{-t},1\rangle. $$

At \(t=0\), since \(e^0=1\),

$$ \mathbf{r}(0)=\langle 1,1,0\rangle, \qquad \mathbf{r}'(0)=\langle 1,-1,1\rangle. $$

Their dot product is \(1(1)+1(-1)+0(1)=1-1+0=0\). Hence

$$ \left.\frac{d}{dt}\lVert\mathbf{r}(t)\rVert^2\right|_{t=0} =2(0)=0. $$

To check, expand the squared magnitude:

$$ \lVert\mathbf{r}(t)\rVert^2=e^{2t}+e^{-2t}+t^2, \qquad \frac{d}{dt}\lVert\mathbf{r}(t)\rVert^2=2e^{2t}-2e^{-2t}+2t. $$

At \(t=0\), this derivative is \(2(1)-2(1)+0=0\), matching the result from the dot product.

A Common Mistake: Forgetting the Two Terms

A frequent error is to differentiate \(\mathbf{r}(t)\cdot\mathbf{r}(t)\) as though only one copy of \(\mathbf{r}(t)\) changes. That loses one of the two equal contributions in the dot-product rule.

For example, if \(\mathbf{r}(t)=\langle t,0\rangle\), then \(\lVert\mathbf{r}(t)\rVert^2=t^2\), so its derivative is \(2t\). The incorrect calculation \(\mathbf{r}(t)\cdot\mathbf{r}'(t)=t\) misses the factor of two. The correction is to apply the product rule to both factors: \(\mathbf{r}'\cdot\mathbf{r}+\mathbf{r}\cdot\mathbf{r}'=t+t=2t\).

Another related distinction is between the squared magnitude and the magnitude itself. The theorem gives the derivative of \(\lVert\mathbf{r}(t)\rVert^2\), not directly the derivative of \(\lVert\mathbf{r}(t)\rVert\). Where \(\lVert\mathbf{r}(t)\rVert\ne0\), differentiating the square-root relation yields

$$ \frac{d}{dt}\lVert\mathbf{r}(t)\rVert = \frac{\mathbf{r}(t)\cdot\mathbf{r}'(t)} {\lVert\mathbf{r}(t)\rVert}. $$

This last expression requires a nonzero magnitude at the parameter value; division by the magnitude is not valid when \(\mathbf{r}(t)=\mathbf{0}\). By contrast, the squared-magnitude formula itself remains valid even when the vector is zero.

Why This Identity Is Useful

Squared magnitudes occur naturally because they avoid a square root and can be easier to differentiate. The identity turns a derivative of a sum of component squares into a single dot product. It also links change in length to the relationship between a vector and its derivative: a zero dot product means the squared length has zero instantaneous rate of change, while a positive or negative dot product indicates increase or decrease at that instant.

These conclusions should be read with care. A zero derivative at one parameter value says only that the squared magnitude has zero instantaneous rate there; it does not by itself prove the magnitude is constant on an interval. The constant-magnitude conclusion follows when the derivative is zero throughout an interval, or when constancy is otherwise known.

Takeaway: Rewrite the squared magnitude as a self-dot-product, apply the dot-product rule to both factors, and use symmetry to obtain \(2\mathbf{r}(t)\cdot\mathbf{r}'(t)\). Check whether the question asks for the squared magnitude or the magnitude itself.

Check Your Understanding

Use the identity and, where appropriate, verify your result by expanding the squared magnitude.

  1. If \(\mathbf{u}(t)=\langle 2t,1-t\rangle\), find \(\frac{d}{dt}\lVert\mathbf{u}(t)\rVert^2\).
  2. For a differentiable vector function \(\mathbf{r}(t)\), what condition on \(\mathbf{r}(t)\cdot\mathbf{r}'(t)\) says that the squared magnitude has zero derivative at a particular parameter value?
  3. Explain why constant magnitude on an interval implies \(\mathbf{r}(t)\cdot\mathbf{r}'(t)=0\) throughout that interval.
  4. Why is it incorrect to use \(\mathbf{r}(t)\cdot\mathbf{r}'(t)\) alone as the derivative of the squared magnitude?
  5. At a parameter value where \(\mathbf{r}(t)=\mathbf{0}\), can the squared-magnitude derivative formula still be used? Explain.

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