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Chain Rule and Implicit Differentiation · Tutorial 358 of 999

Explicit Time Dependence and Moving Inputs

Combine explicit time dependence with changes in spatial inputs to find and interpret a quantity’s total rate of change.

Intermediate 9 min read

What You'll Learn

  • Distinguish a partial derivative with respect to time from a total time derivative along a path
  • Apply the chain rule to a function of two moving spatial inputs and time
  • Identify the assumptions needed for the total-time-derivative formula
  • Compute total rates of change and check them by direct substitution
  • Avoid omitting explicit time dependence or counting it twice

Introduction

A quantity can change over time for more than one reason. Its spatial location may move through a field, and the field itself may also change with time. To find the quantity’s actual rate of change, both sources must be included.

For instance, let \(T(x,y,t)\) describe temperature at location \((x,y)\) and time \(t\). A moving object follows the path \((x(t),y(t))\), so the temperature it experiences is the single-variable function \(T(x(t),y(t),t)\). Differentiating this quantity requires the chain rule, including the direct dependence on \(t\).

Central distinction: \(T_t\) measures the change in \(T\) with time while \(x\) and \(y\) are held fixed. The total derivative along a moving path also includes the effects of changes in \(x\) and \(y\).

The Total Time Derivative

Suppose \(T\) is a differentiable function of three variables, \(T=T(x,y,t)\), and suppose \(x=x(t)\) and \(y=y(t)\) are differentiable paths. Define the quantity along the path by \(F(t)=T(x(t),y(t),t)\). The derivative \(dF/dt\) is the total time derivative: it accounts for every way the value can change as time advances.

Definition: For a differentiable function \(T(x,y,t)\) evaluated along differentiable paths \(x(t)\) and \(y(t)\), its total time derivative is
$$ \frac{d}{dt}T(x(t),y(t),t) = T_x(x(t),y(t),t)\,x'(t) + T_y(x(t),y(t),t)\,y'(t) + T_t(x(t),y(t),t). $$
Here \(T_x\), \(T_y\), and \(T_t\) are partial derivatives. In \(T_t\), the spatial inputs are held fixed; in \(T_x\) and \(T_y\), the other inputs, including \(t\), are held fixed.

The first two terms describe changes caused by motion through the spatial inputs. The last term describes the change that would occur from time passing even if the spatial location were fixed. The values of the partial derivatives and the input rates are all evaluated at the same time and along the same path.

In particular, \(T_t\) is not generally the same as \(dT/dt\). It is only one contribution to the total derivative. If \(T\) has no explicit dependence on time, then \(T_t=0\), and only the spatial-motion terms remain. If the spatial inputs are momentarily stationary, then \(x'=y'=0\), but the total derivative can still be nonzero when \(T_t\ne 0\).

Why the Formula Works

The formula is an application of the multivariable chain rule along a parametrized path. A short derivation also clarifies why the three contributions are added rather than treated as alternatives.

Fix a time \(t\), and let \(h\) be a small time increment. The inputs change by \(\Delta x=x(t+h)-x(t)\), \(\Delta y=y(t+h)-y(t)\), and \(\Delta t=h\). Since \(T\) is differentiable, its first-order change is given by its linear approximation at \((x(t),y(t),t)\), with a remainder smaller than the size of the input change.

$$ \begin{aligned} &T(x(t+h),y(t+h),t+h)-T(x(t),y(t),t)\\ &\quad=T_x\,\Delta x+T_y\,\Delta y+T_t\,h+o\!\left(\sqrt{(\Delta x)^2+(\Delta y)^2+h^2}\right). \end{aligned} $$

Because \(x\) and \(y\) are differentiable at \(t\), their changes satisfy \(\Delta x=x'(t)h+o(h)\) and \(\Delta y=y'(t)h+o(h)\). Thus the size of the combined input change is bounded by a constant times \(|h|\) for sufficiently small \(h\), and the remainder divided by \(h\) tends to zero. Divide the displayed equation by \(h\) and let \(h\) tend to zero. The left side becomes the derivative of the composition, while the three linear terms become \(T_xx'(t)\), \(T_yy'(t)\), and \(T_t\). This gives the stated formula.

Conditions: The formula applies at a time \(t\) where \(x(t)\) and \(y(t)\) are differentiable and \(T\) is differentiable at the corresponding point \((x(t),y(t),t)\). The partial derivatives and path rates must exist there.

Worked Example: A Polynomial with Explicit Time Dependence

Worked Example: A Polynomial with Explicit Time Dependence

Let \(T(x,y,t)=x^2y+3t\), with \(x(t)=1+t\) and \(y(t)=2-t\). Find the total derivative at \(t=1\).

First calculate the partial derivatives, holding the other inputs fixed:

$$ T_x=2xy,\qquad T_y=x^2,\qquad T_t=3. $$

The path rates are \(x'(t)=1\) and \(y'(t)=-1\). At \(t=1\), the inputs and rates are

$$ x(1)=2,\qquad y(1)=1,\qquad x'(1)=1,\qquad y'(1)=-1. $$

Evaluate the partial derivatives at \((2,1,1)\): \(T_x=2(2)(1)=4\), \(T_y=2^2=4\), and \(T_t=3\). Therefore,

$$ \frac{dT}{dt} = (4)(1)+(4)(-1)+3 = 4-4+3 = 3. $$

Check by substituting the path into \(T\) first:

$$ \begin{aligned} T(x(t),y(t),t) &=(1+t)^2(2-t)+3t\\ &=(1+2t+t^2)(2-t)+3t\\ &=2+3t-t^3+3t\\ &=2+6t-t^3. \end{aligned} $$

Its derivative is \(6-3t^2\), which at \(t=1\) equals \(6-3=3\). This agrees with the chain-rule calculation.

Separating Spatial Change from Explicit Time Change

The three terms in the formula have distinct meanings. The spatial terms measure how the value changes as the point moves through the field at the current time. The explicit time term measures how the field changes at a fixed point. A moving path can make the spatial contributions cancel even while the explicit time contribution remains.

Worked Example: Motion Along a Circle in a Changing Field

Let \(T(x,y,t)=e^{-t}(x^2+y^2)\), with \(x(t)=3\cos t\) and \(y(t)=3\sin t\). Find \(dT/dt\) at \(t=0\), and interpret the contributions.

Differentiate with respect to each input while holding the others fixed:

$$ T_x=2xe^{-t},\qquad T_y=2ye^{-t},\qquad T_t=-(x^2+y^2)e^{-t}. $$

The path values at \(t=0\) are \(x(0)=3\) and \(y(0)=0\). The rates are \(x'(t)=-3\sin t\) and \(y'(t)=3\cos t\), so \(x'(0)=0\) and \(y'(0)=3\). At \((3,0,0)\), the partial derivatives are \(T_x=6\), \(T_y=0\), and \(T_t=-9\). Thus,

$$ \frac{dT}{dt} = (6)(0)+(0)(3)+(-9) = -9. $$

To check, note that \(x(t)^2+y(t)^2=9\cos^2t+9\sin^2t=9\). Hence \(T(x(t),y(t),t)=9e^{-t}\), whose derivative is \(-9e^{-t}\), equal to \(-9\) at \(t=0\). The spatial contributions sum to zero because the path stays at a constant distance from the origin; the field’s explicit time dependence accounts for the change.

Worked Example: A Moving Location and a Changing Formula

Worked Example: A Moving Location and a Changing Formula

Let \(T(x,y,t)=xy+tx\), with \(x(t)=t^2\) and \(y(t)=1+2t\). Find the total derivative at \(t=1\), showing each contribution separately.

The partial derivatives are

$$ T_x=y+t,\qquad T_y=x,\qquad T_t=x. $$

The path rates are \(x'(t)=2t\) and \(y'(t)=2\). At \(t=1\), \(x=1\), \(y=3\), \(x'=2\), and \(y'=2\). Therefore \(T_x=3+1=4\), \(T_y=1\), and \(T_t=1\). The total derivative is

$$ \frac{dT}{dt} = (4)(2)+(1)(2)+1 = 8+2+1 = 11. $$

For a direct check, substitute the paths before differentiating:

$$ \begin{aligned} T(x(t),y(t),t) &=t^2(1+2t)+t(t^2)\\ &=t^2+2t^3+t^3\\ &=t^2+3t^3. \end{aligned} $$

Differentiating gives \(2t+9t^2\), whose value at \(t=1\) is \(2+9=11\). The direct calculation confirms the total derivative, including both spatial motion and explicit time dependence.

Interpreting Units and Signs

The formula also provides a useful consistency check. Suppose \(T\) is measured in degrees, \(x\) and \(y\) in metres, and \(t\) in seconds. Then \(T_x\) and \(T_y\) have units of degrees per metre, while \(x'\) and \(y'\) have units of metres per second. Consequently, \(T_xx'\) and \(T_yy'\) have units of degrees per second. The partial derivative \(T_t\) also has units of degrees per second, so every term can be added.

The signs of the terms describe whether each source increases or decreases the quantity at that instant. A negative contribution does not guarantee the total derivative is negative: positive contributions from other terms may outweigh it. The formula adds the signed contributions, not their absolute values.

Takeaway: The total rate is the sum of the field’s explicit time-change and the changes induced by every moving input. Each term has the same units as the total rate.

Common Mistake: Omitting the Explicit Time Term

A frequent error is to apply the path terms but forget \(T_t\). For a concrete example, let \(T(x,y,t)=x^2+y^2+t\), and keep the location fixed at \(x(t)=1\), \(y(t)=2\). Then \(x'=y'=0\), and the partial derivatives with respect to the spatial inputs contribute nothing. If one incorrectly writes only \(T_xx'+T_yy'\), the result is \(0\).

But \(T_t=1\), so the correct calculation is

$$ \frac{dT}{dt} = T_xx'+T_yy'+T_t = (2x)(0)+(2y)(0)+1 = 1. $$

Indeed, substitution gives \(T(1,2,t)=1+4+t=5+t\), whose derivative is \(1\). The location is fixed, but the quantity still changes because the formula itself depends on time.

The opposite bookkeeping error is to treat \(T_t\) as though it already included the effects of changing \(x\) and \(y\). It does not: its definition holds those inputs fixed. Include the spatial terms separately, and include \(T_t\) exactly once.

Check Your Understanding

Use the total-time-derivative formula to answer the following questions. When a path and a time are given, evaluate every term at that time.

  1. For \(T(x,y,t)=x+y+t\), with \(x(t)=t^2\) and \(y(t)=3t\), find \(dT/dt\) at \(t=2\).
  2. In the formula for \(dT/dt\), which inputs are held fixed when computing \(T_t\)?
  3. If \(x'(t)=y'(t)=0\) at a particular instant, what does the total derivative reduce to?
  4. For \(T(x,y,t)=x^2+y^2+t\), with \(x(t)=\cos t\) and \(y(t)=\sin t\), compute \(dT/dt\) and explain which contribution remains.
  5. State the differentiability conditions needed to apply the total-time-derivative formula at a given time.

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