Wemazi

Topic 9.2 · BC

Convert to the second derivative with respect to x

Builds on differentiating the slope with respect to the parameter. 7 minutes.

From a rate per t to a rate per x

A point moves with \(x = t^2\) and \(y = \frac{t^4}{4}\), for \(t > 0\). The slope of its path is \(m = \frac{t^2}{2}\), and the lower graph shows \(m\) against \(x\).

Move \(t\) with the slider. The triangle at the orange point on the lower graph shows the changes during one unit of \(t\). Its horizontal leg is \(\frac{dx}{dt}\), the change in \(x\), and its vertical leg is \(\frac{dm}{dt}\), the change in \(m\). Both legs are green because both changes are positive.

At \(t = 1\), the legs are 2 and 1, so the slope of the lower graph is \(\frac12\). At \(t = 2\), the legs are 4 and 2, and the slope is again \(\frac12\).

The lower graph is a straight line, because the path is the parabola \(y = \frac{x^2}{4}\), so \(m = \frac{x}{2}\). Its slope is the change in \(m\) for one unit of \(x\), which is the second derivative of \(y\) with respect to \(x\).

The ratio does not depend on the speed of t

The lower graph again shows \(m\) against \(x\), and its triangle has a horizontal leg \(\frac{dx}{dt}\) and a vertical leg \(\frac{dm}{dt}\).

Both legs are changes during one unit of \(t\), so they depend on how fast \(t\) runs. Double the speed. Both legs double, and the orange line does not turn.

The slope of the orange line is the change in \(m\) per unit of \(x\), so it does not depend on the speed of \(t\). It equals \(\frac{dm}{dt}\) divided by \(\frac{dx}{dt}\).

Notation

  1. Find \(\frac{dm}{dt}\), the derivative of the slope \(m = \frac{dy}{dx}\) with respect to \(t\), as in the last skill. The lower graph shows \(m\) against \(x\).

  2. The second derivative is the derivative of \(m\) with respect to \(x\). To change a rate per unit of \(t\) into a rate per unit of \(x\), divide by \(\frac{dx}{dt}\).

    \[\cy{\dfrac{d^2y}{dx^2}} = \dfrac{\cg{dm/dt}}{\cg{dx/dt}} = \dfrac{\dfrac{d}{dt}\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}\]

  3. This is the rule that gave \(\frac{dy}{dx}\). The rate of any quantity with respect to \(x\) is its rate with respect to \(t\), divided by \(\frac{dx}{dt}\).

  4. Divide by the first derivative \(\frac{dx}{dt}\), and not by the second derivative \(\frac{d^2x}{dt^2}\). The fraction \(\frac{d^2y/dt^2}{d^2x/dt^2}\) is a different quantity.

  5. The rate \(\frac{dx}{dt}\) must not be 0, because the formula divides by it.

When it fails

On the curve \(x = t^3\), \(y = t^4\), the second derivative is \(\frac{4}{9t^2}\). The formula divides by \(\frac{dx}{dt} = 3t^2\), which is 0 at \(t = 0\).

Move \(t\) toward 0. The horizontal leg shrinks to 0, and the orange line on the lower graph turns vertical. At \(t = 0\) the formula has no value, and near it the ratio \(\frac{4}{9t^2}\) is very large.

Example 1: Exponential rates

For \(x = e^{2t}\) and \(y = e^{-t}\), find \(\frac{d^2y}{dx^2}\) as a function of \(t\).

  1. Find the rates and the slope.

    \[\begin{aligned} \dfrac{dx}{dt} &= 2e^{2t},\quad \dfrac{dy}{dt} = -e^{-t} \\ m &= \dfrac{-e^{-t}}{2e^{2t}} = -\dfrac12e^{-3t} \end{aligned}\]

  2. Differentiate \(m\) with respect to \(t\). At \(t = 0\) this is the vertical leg, \(\frac32\).

    \[\dfrac{dm}{dt} = \dfrac32e^{-3t}\]

  3. Divide by \(\frac{dx}{dt} = 2e^{2t}\), and subtract the exponents of \(e\). At \(t = 0\) the horizontal leg is 2, and the slope of the orange line is \(\frac34\).

    \[\cy{\dfrac{d^2y}{dx^2}} = \dfrac{\frac32e^{-3t}}{2e^{2t}} = \dfrac34e^{-5t}\]

  4. To test the result, write \(y = x^{-1/2}\). Then \(\frac{d^2y}{dx^2} = \frac34x^{-5/2}\). Because \(x = e^{2t}\), this is \(\frac34e^{-5t}\), the same result.

Example 2: A value at one point

For \(x = t + \ln t\) and \(y = t^2\), with \(t > 0\), find \(\frac{d^2y}{dx^2}\) at \(t = 1\).

  1. Find the rates and the slope.

    \[\begin{aligned} \dfrac{dx}{dt} &= 1 + \dfrac1t = \dfrac{t + 1}{t} \\ \dfrac{dy}{dt} &= 2t,\quad m = \dfrac{2t^2}{t + 1} \end{aligned}\]

  2. Differentiate \(m\) with respect to \(t\) with the quotient rule. At \(t = 1\) the result is \(\frac32\), the vertical leg.

    \[\begin{aligned} \dfrac{dm}{dt} &= \dfrac{4t(t + 1) - 2t^2}{(t + 1)^2} \\ &= \dfrac{2t^2 + 4t}{(t + 1)^2} \\ \dfrac{dm}{dt}\Big|_{t=1} &= \dfrac32 \end{aligned}\]

  3. A tempting wrong move: The second derivative is the derivative of the slope, so the answer is \(\frac32\).

    \[\dfrac{d^2y}{dx^2}\ \overset{?}{=}\ \cr{\dfrac32}\]

    The value \(\frac32\) is the change in \(m\) for one unit of \(t\). In one unit of \(t\), \(x\) changes by 2, so \(m\) changes less per unit of \(x\). The red line with slope \(\frac32\) is steeper than the orange side of the triangle.

  4. Divide by \(\frac{dx}{dt}\), which is 2 at \(t = 1\).

    \[\cy{\dfrac{d^2y}{dx^2}} = \dfrac{3/2}{2} = \dfrac34\]

Practice

For \(x = \dfrac1t\) and \(y = \ln t\), with \(t > 0\), find \(\frac{d^2y}{dx^2}\) as a function of \(t\). Then find its value at \(t = 3\).

Find the slope \(m\) first. Differentiate \(m\) with respect to \(t\). Then divide by \(\frac{dx}{dt}\). On the lower graph, the answer is the slope of the graph of \(m\) where \(t = 3\).

Solution

  1. Find the rates and the slope.

    \[\begin{aligned} \dfrac{dx}{dt} &= -\dfrac{1}{t^2},\quad \dfrac{dy}{dt} = \dfrac1t \\ m &= \dfrac{1/t}{-1/t^2} = -t \end{aligned}\]

  2. Differentiate \(m\) with respect to \(t\). Both legs of the triangle are red, because \(x\) and \(m\) both decrease while \(t\) increases.

    \[\dfrac{dm}{dt} = -1\]

  3. Divide by \(\frac{dx}{dt}\). Then substitute \(t = 3\).

    \[\begin{aligned} \cy{\dfrac{d^2y}{dx^2}} &= \dfrac{-1}{-1/t^2} = t^2 \\ \cy{\dfrac{d^2y}{dx^2}}\Big|_{t=3} &= 9 \end{aligned}\]

Questions

The upper graph shows the curve \(x = 2t\), \(y = t^2\). The lower graph shows its slope \(m\) against \(x\). The triangle at the orange point on the lower graph shows the two changes during one unit of \(t\).

  1. What is \(\frac{d^2y}{dx^2}\) at the orange point?

    • \(\frac12\)
    • 1
    • 2
    Answer

    \(\frac12\)

  2. The orange point moves to \(t = 2\). How does \(\frac{d^2y}{dx^2}\) change?

    • It does not change
    • It doubles
    • It becomes 0
    Answer

    It does not change

  3. For \(x = t^2 + 4\) and \(y = t^3\), what is \(\frac{d^2y}{dx^2}\) at \(t = 3\)?

    • \(\frac14\)
    • \(\frac32\)
    • 9
    Answer

    \(\frac14\)

Summary

You differentiate the slope with respect to \(t\), and then divide by \(\frac{dx}{dt}\).

It works because \(\frac{dm}{dt}\) and \(\frac{dx}{dt}\) are changes during one unit of \(t\), so their ratio is the change in slope per unit of \(x\).

You use it to find \(\frac{d^2y}{dx^2}\) for a curve given by \(x = f(t)\) and \(y = g(t)\), where \(\frac{dx}{dt}\) is not 0.

This is one of four skills in topic 9.2 of Which & Why: AP Calculus.