Differentiate the slope with respect to the parameter
How fast the slope changes with t
A point moves along the path \(x = 2t\), \(y = t^3\) in the upper graph. Its slope is \(m = \frac{dy/dt}{dx/dt} = \frac{3t^2}{2}\), and the lower graph shows \(m\) against \(t\).
At \(t = 1\), the slope of the path is \(\frac32\), and the orange line that touches the lower graph there has slope 3. This slope is the rate of change of \(m\) with respect to \(t\). Near \(t = 1\), the slope of the path rises by about 3 for each unit of \(t\).
At \(t = 0\), the orange line is flat. The graph of \(m\) is at its lowest point there, so the slope of the path is not changing for an instant.
A derivative with respect to t
The lower graph shows the slope \(m\) of the path against \(t\). The orange line touches it at the current \(t\), and its slope \(\cy{\frac{dm}{dt}}\) tells how fast the slope of the path changes as \(t\) increases.
Because the lower graph has \(t\) on its horizontal axis, \(\frac{dm}{dt}\) is a derivative with respect to \(t\), not yet with respect to \(x\).
Notation
Write the slope as a function of \(t\), and call it \(m\). It is a fraction, with the rate of \(y\) on top and the rate of \(x\) on the bottom.
\[\begin{aligned} m(t) &= \dfrac{dy}{dx} = \dfrac{u}{v} \\ u &= \dfrac{dy}{dt},\quad v = \dfrac{dx}{dt} \end{aligned}\]Differentiate \(m\) with respect to \(t\) with the quotient rule, because \(m\) is a fraction. The orange line on the lower graph has this slope.
\[\cy{\dfrac{dm}{dt}} = \dfrac{u'v - uv'}{v^2}\]If \(\frac{dx}{dt}\) is a constant, then \(v' = 0\) and the bottom stays the same, so you differentiate only the top.
\[\dfrac{dm}{dt} = \dfrac{u'}{v}\]Because \(\frac{dm}{dt}\) is the derivative of a fraction, it is not the ratio of \(\frac{d^2y}{dt^2}\) to \(\frac{d^2x}{dt^2}\).
The functions \(x\) and \(y\) must be twice differentiable in \(t\), and \(\frac{dx}{dt}\) must not be 0.
When it fails
On the curve \(x = 2t^3\), \(y = t\), the slope is \(m = \frac{1}{6t^2}\), and its bottom \(\frac{dx}{dt} = 6t^2\) is 0 at \(t = 0\).
Move t to 0. The slope has no value there, so the lower graph breaks and the orange line disappears. The tangent to the path is vertical, and there is no slope to differentiate.
Example 1: A constant horizontal rate
For \(x = 4t\) and \(y = 2t^3 - 3t\), let \(m = \frac{dy}{dx}\). Find \(\frac{dm}{dt}\), and then evaluate it at \(t = 1\).
Find the rates. The horizontal rate is a constant.
\[\dfrac{dx}{dt} = 4,\qquad \dfrac{dy}{dt} = 6t^2 - 3\]Write the slope as a function of \(t\).
\[m = \dfrac{6t^2 - 3}{4}\]Differentiate with respect to \(t\). The bottom is the constant 4, so you differentiate only the top.
\[\cy{\dfrac{dm}{dt}} = \dfrac{12t}{4} = 3t\]Substitute \(t = 1\). The orange line on the lower graph has slope 3. Near this point, the slope of the path rises by about 3 for each unit of \(t\).
\[\cy{\dfrac{dm}{dt}}\Big|_{t=1} = 3\]
Example 2: A slope that is a fraction
For \(x = t^2 + t\) and \(y = t^3\), let \(m = \frac{dy}{dx}\). Find \(\frac{dm}{dt}\) at \(t = 1\).
Find the rates, and write the slope as a function of t.
\[\begin{aligned} \dfrac{dx}{dt} &= 2t + 1,\quad \dfrac{dy}{dt} = 3t^2 \\ m &= \dfrac{3t^2}{2t + 1} \end{aligned}\]A tempting wrong move: Differentiate the top and the bottom separately. The result is \(\frac{d^2y}{dt^2}\) divided by \(\frac{d^2x}{dt^2}\).
\[\dfrac{dm}{dt}\ \overset{?}{=}\ \dfrac{6t}{2} = \cr{3t}\]At \(t = 1\), this gives 3. The red line with slope 3 climbs much faster than the graph of \(m\). The derivative of a fraction is not the fraction of the derivatives.
Use the quotient rule with \(u = 3t^2\) and \(v = 2t + 1\).
\[\begin{aligned} \cy{\dfrac{dm}{dt}} &= \dfrac{6t(2t + 1) - 3t^2\cdot 2}{(2t + 1)^2} \\ &= \dfrac{6t^2 + 6t}{(2t + 1)^2} \end{aligned}\]Substitute \(t = 1\). The orange line touches the graph of \(m\) and has slope \(\frac43\).
\[\cy{\dfrac{dm}{dt}}\Big|_{t=1} = \dfrac{12}{9} = \dfrac43\]
Practice
For \(x = 2t + \sin t\) and \(y = \cos t\), let \(m = \frac{dy}{dx}\). Find \(\frac{dm}{dt}\) at \(t = 0\).
Use the quotient rule on \(m\). Then use \(\cos^2 t + \sin^2 t = 1\) to simplify the top. The answer is the slope of the lower graph at \(t = 0\).
Solution
Find the rates and write the slope.
\[\begin{aligned} \dfrac{dx}{dt} &= 2 + \cos t,\quad \dfrac{dy}{dt} = -\sin t \\ m &= \dfrac{-\sin t}{2 + \cos t} \end{aligned}\]Use the quotient rule.
\[\dfrac{dm}{dt} = \dfrac{-\cos t\,(2 + \cos t) - \sin^2 t}{(2 + \cos t)^2}\]The top is \(-2\cos t - \cos^2 t - \sin^2 t\). Because \(\cos^2 t + \sin^2 t = 1\), it simplifies.
\[\dfrac{dm}{dt} = \dfrac{-(2\cos t + 1)}{(2 + \cos t)^2}\]Substitute \(t = 0\). The orange line on the lower graph slopes down.
\[\cy{\dfrac{dm}{dt}}\Big|_{t=0} = \dfrac{-3}{9} = -\dfrac13\]
Questions
The upper graph shows the curve \(x = 4t\), \(y = t^4 - 2t^2\). The lower graph shows its slope \(m\) against \(t\). The orange line touches the lower graph at the same value of \(t\) as the orange point.
At the orange point, where t = 0, what is the slope of the path doing as t increases?
- It is decreasing
- It is increasing
- It is not changing
Answer
It is decreasing
For \(t > 0\), at which value of \(t\) is \(\frac{dm}{dt} = 0\)?
- \(t = \frac{1}{\sqrt3} \approx 0.58\)
- \(t = 1\)
- \(t = 0\)
Answer
\(t = \frac{1}{\sqrt3} \approx 0.58\)
For \(x = t^3\) and \(y = t^2\), let \(m = \frac{dy}{dx}\). What is \(\frac{dm}{dt}\) at \(t = 1\)?
- \(-\frac23\)
- \(\frac13\)
- \(\frac23\)
Answer
\(-\frac23\)
Summary
You write the slope \(m\) as a fraction in \(t\) and differentiate it, with the quotient rule if the bottom is not a constant.
This works because \(m\) is a function of the parameter, so its rate of change is taken with respect to \(t\).
The result is a rate per unit of \(t\), not yet per unit of \(x\), so it is the middle step toward the second derivative.
This is one of four skills in topic 9.2 of Which & Why: AP Calculus.