From a Scalar Field to a Rate Along a Path
A scalar field assigns a number to each point in a region. For example, a field might describe temperature at each location, while a parametrized path describes the locations visited by a moving object. Combining them gives a function of the path parameter: at each parameter value, the composite function records the field value at the current position.
The central question is how quickly that field value changes as the parameter changes. The answer uses two pieces of information: the gradient of the field at the current point, and the velocity of the path at that parameter value.
Definition and Conditions
Let \(U\) be an open subset of \(\mathbb R^n\), let \(f:U\to\mathbb R\) be differentiable, and let \(\mathbf r:I\to U\) be a differentiable path on an interval \(I\). The scalar-valued function obtained by evaluating the field along the path is \(F(t)=f(\mathbf r(t))\). Differentiating \(F\) measures the rate at which the field value changes with respect to the parameter \(t\).
In coordinates, if \(\mathbf r(t)=(x_1(t),\ldots,x_n(t))\), then the same rule is $$ \frac{d}{dt}f(\mathbf r(t)) = \sum_{i=1}^{n} \frac{\partial f}{\partial x_i}(\mathbf r(t))\,x_i'(t). $$ Each partial derivative measures the field’s change in one coordinate direction, and the corresponding coordinate velocity tells how quickly the path moves in that direction. The sum combines all coordinate contributions.
Why the Formula Works
The gradient gives the linear approximation to the change in a differentiable scalar field. For a small displacement \(\mathbf h\) from a point \(\mathbf a\), differentiability means $$ f(\mathbf a+\mathbf h) = f(\mathbf a)+\nabla f(\mathbf a)\cdot\mathbf h +o(\|\mathbf h\|). $$ The final term is an error that becomes negligible compared with \(\|\mathbf h\|\) as \(\mathbf h\) approaches the zero vector.
To apply this to a path at \(t_0\), set \(\mathbf a=\mathbf r(t_0)\) and consider the path displacement \(\mathbf h=\mathbf r(t_0+s)-\mathbf r(t_0)\), where \(s\) is a small change in parameter. Since the path is differentiable, $$ \mathbf h=s\mathbf r'(t_0)+o(|s|). $$ Substituting this displacement into the field’s linear approximation gives the change in the composite function. Dividing by \(s\), the linear term tends to \(\nabla f(\mathbf r(t_0))\cdot\mathbf r'(t_0)\). The error term divided by \(s\) tends to zero: the path displacement is \(O(|s|)\), and the differentiability remainder is \(o(\|\mathbf h\|)\). Thus the difference quotient has the stated limit.
This argument explains the roles of the two vectors. The gradient describes how the field responds to a small displacement; the velocity describes the displacement produced by a small change in parameter.
Worked Examples
Worked Example: A Path in Two Variables
Let \(f(x,y)=x^2+3y\) and \(\mathbf r(t)=(t^2,2t+1)\). Find the rate of change of the field along the path at \(t=1\), then verify it by differentiating the composite function.
First calculate the gradient: $$ \nabla f(x,y)=(2x,3). $$ At \(t=1\), the path is $$ \mathbf r(1)=(1^2,2(1)+1)=(1,3), $$ so $$ \nabla f(\mathbf r(1))=(2(1),3)=(2,3). $$ The path velocity is $$ \mathbf r'(t)=(2t,2), \qquad \mathbf r'(1)=(2,2). $$ Therefore the rate is $$ \nabla f(\mathbf r(1))\cdot\mathbf r'(1) =(2,3)\cdot(2,2) =2(2)+3(2)=10. $$
To check, compose the functions directly: $$ f(\mathbf r(t))=(t^2)^2+3(2t+1)=t^4+6t+3. $$ Differentiating gives $$ \frac{d}{dt}f(\mathbf r(t))=4t^3+6, \qquad \left.\frac{d}{dt}f(\mathbf r(t))\right|_{t=1} =4(1)^3+6=10. $$ The direct calculation agrees with the gradient-velocity calculation.
Worked Example: A Path in Three Variables
Let \(f(x,y,z)=xy+z^2\) and \(\mathbf r(t)=(\cos t,\sin t,t)\). Find the rate along the path at \(t=\frac{\pi}{4}\) and check it by differentiating the composite function.
The gradient and path velocity are $$ \nabla f(x,y,z)=(y,x,2z), \qquad \mathbf r'(t)=(-\sin t,\cos t,1). $$ At \(t=\frac{\pi}{4}\), the path point is $$ \mathbf r\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},\frac{\pi}{4}\right). $$ Thus $$ \nabla f\left(\mathbf r\left(\frac{\pi}{4}\right)\right) = \left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},\frac{\pi}{2}\right), $$ and $$ \mathbf r'\left(\frac{\pi}{4}\right) = \left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},1\right). $$ Their dot product is $$ -\frac{1}{2}+\frac{1}{2}+\frac{\pi}{2} =\frac{\pi}{2}. $$
For the check, evaluate the field along the path: $$ f(\mathbf r(t))=(\cos t)(\sin t)+t^2. $$ Its derivative is $$ \frac{d}{dt}f(\mathbf r(t)) =\cos^2t-\sin^2t+2t. $$ At \(t=\frac{\pi}{4}\), the squared trigonometric terms are both \(\frac12\), so $$ \cos^2\left(\frac{\pi}{4}\right) -\sin^2\left(\frac{\pi}{4}\right) +2\left(\frac{\pi}{4}\right) =0+\frac{\pi}{2} =\frac{\pi}{2}. $$ This verifies the rate.
Worked Example: An Exponential Field Along a Curved Path
Let \(f(x,y)=e^{x+y}\) and \(\mathbf r(t)=(t,t^2)\). Find the rate of change at \(t=1\), and verify the answer directly.
The gradient is $$ \nabla f(x,y)=(e^{x+y},e^{x+y}). $$ At the path point \(\mathbf r(1)=(1,1)\), this gives $$ \nabla f(\mathbf r(1))=(e^2,e^2). $$ The velocity is $$ \mathbf r'(t)=(1,2t), \qquad \mathbf r'(1)=(1,2). $$ Therefore the rate at \(t=1\) is $$ (e^2,e^2)\cdot(1,2)=e^2+2e^2=3e^2. $$
Direct composition gives $$ f(\mathbf r(t))=e^{t+t^2}. $$ Differentiating, $$ \frac{d}{dt}f(\mathbf r(t)) =e^{t+t^2}(1+2t). $$ At \(t=1\), this is $$ e^{1+1}(1+2)=3e^2, $$ which matches the result from the gradient and velocity.
Parameter Speed and Directional Derivatives
The formula gives a rate per unit change in the parameter \(t\). It does not automatically give a rate per unit distance traveled. When \(\mathbf r'(t)\ne\mathbf 0\), define the unit tangent vector $$ \mathbf u(t)=\frac{\mathbf r'(t)}{\|\mathbf r'(t)\|}. $$ Then the chain rule can also be written as $$ \frac{d}{dt}f(\mathbf r(t)) = \|\mathbf r'(t)\|\, \bigl(\nabla f(\mathbf r(t))\cdot\mathbf u(t)\bigr). $$ The dot product in parentheses is the directional derivative in the unit tangent direction. The factor \(\|\mathbf r'(t)\|\) is the path’s speed. Thus the rate with respect to the parameter is the rate per unit distance multiplied by the distance traveled per unit parameter.
If \(\mathbf r'(t)=\mathbf 0\), the unit tangent is not defined, so this decomposition cannot be used at that parameter. The original chain rule still applies whenever \(f\) and \(\mathbf r\) meet the stated differentiability conditions.
A Common Mistake: Treating Velocity as a Unit Direction
Consider \(f(x,y)=x^2+y^2\) along \(\mathbf r(t)=(3t,4t)\). At \(t=1\), the point is \((3,4)\), the gradient is \((6,8)\), and the velocity is \((3,4)\). The correct rate is $$ (6,8)\cdot(3,4)=6(3)+8(4)=18+32=50. $$
A common incorrect calculation normalizes the velocity and then reports the directional derivative as the rate along the parameter: $$ \mathbf u=\left(\frac35,\frac45\right), \qquad (6,8)\cdot\mathbf u =\frac{18}{5}+\frac{32}{5} =10. $$ This value is the rate per unit distance, not per unit \(t\). The path has speed $$ \|\mathbf r'(1)\|=\sqrt{3^2+4^2}=5, $$ so the rate per unit parameter is \(10(5)=50\). Checking directly, \(f(\mathbf r(t))=(3t)^2+(4t)^2=25t^2\), whose derivative at \(t=1\) is \(50\). The distinction is essential whenever the path’s speed is not one.
Using the Formula Reliably
Write \(f\) and \(\mathbf r(t)\), and confirm that the field is defined at the path points under consideration.
Compute \(\nabla f\) as a function of position and differentiate each component of \(\mathbf r(t)\).
Substitute \(\mathbf r(t)\) into the gradient, then take its dot product with \(\mathbf r'(t)\).
When practical, form \(f(\mathbf r(t))\), differentiate it directly, and compare the values.
Keeping the evaluation points aligned prevents a frequent source of error: the gradient must be evaluated at the current path point, not at an unrelated location. Also preserve the path’s actual velocity rather than replacing it with a unit vector unless the question asks specifically for a rate per unit distance.
Check Your Understanding
Use the chain rule along paths to answer each question. Where requested, verify by differentiating the composite function.
- For \(f(x,y)=x+2y\) and \(\mathbf r(t)=(t^2,3t)\), find the rate at \(t=2\).
- For \(f(x,y)=xy\) and \(\mathbf r(t)=(t,1-t)\), calculate \(\frac{d}{dt}f(\mathbf r(t))\).
- State the differentiability conditions needed to apply the rate formula at a parameter value.
- Explain why a directional derivative in a unit tangent direction may differ from the rate with respect to \(t\).
- If \(\mathbf r'(t_0)=\mathbf 0\) and \(f\) is differentiable at \(\mathbf r(t_0)\), what does the chain rule imply for the rate at \(t_0\)?