Wemazi

Topic 9.1 · BC

Calculate the slope of a parametric curve

Builds on differentiating with the chain rule. 7 minutes.

A moving point and its slope

A point moves in the plane, and at time \(t\) it is at \(x = 2t\), \(y = 6t - t^2\). Drag the orange point along the curve, or move the slider.

The small triangle shows how far the point would move in one unit of time if it kept its present speed and direction. Its horizontal leg is the change in \(x\). Its vertical leg is the change in \(y\), green when \(y\) increases and red when \(y\) decreases.

At \(t = 1\), the point moves 2 across and 4 up. The orange line along this motion is the tangent line, so its slope is 4 divided by 2, which is 2.

At \(t = 5\), the point moves 2 across and 4 down, so the vertical leg is red and the slope is \(-2\).

The slope is a ratio of two rates

The curve now stands for any curve \(x = f(t)\), \(y = g(t)\). Its legs are \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), the rates at which \(x\) and \(y\) change per unit of \(t\).

Move the speed slider. Both legs change by the same factor, because the point moves faster or slower in both directions, so the orange line does not turn. The faint triangle shows speed 1.

The direction of the line depends only on the ratio of the two legs, and this ratio is the slope.

Notation

  1. The two legs are the derivatives of \(x\) and of \(y\) with respect to \(t\), the rates at which the point moves across and up.

    \[\cg{\dfrac{dx}{dt}},\qquad \cg{\dfrac{dy}{dt}}\]

  2. The slope of a line is its rise divided by its run. Here the rise is \(\frac{dy}{dt}\) and the run is \(\frac{dx}{dt}\), so the tangent line has this slope.

    \[\cy{\dfrac{dy}{dx}} = \dfrac{\cg{dy/dt}}{\cg{dx/dt}}\]

  3. The chain rule gives the same result, because \(y\) depends on \(x\) and \(x\) depends on \(t\). Divide both sides by \(\frac{dx}{dt}\) to find the slope.

    \[\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt}\]

  4. In words, the slope is the rate of \(y\) divided by the rate of \(x\), both with respect to \(t\).

  5. Two conditions must hold. Both \(x\) and \(y\) must be differentiable in \(t\), and \(\frac{dx}{dt}\) must not be 0, because the formula divides by it.

    Each value of \(t\) gives a number for the slope, so you do not need an equation in \(x\) and \(y\).

When it fails

Here \(x = (t-2)^2\) and \(y = t\), so \(\frac{dx}{dt} = 2(t-2)\) is 0 at \(t = 2\). Move the point there. The horizontal leg has length 0, so the point moves straight up.

The formula divides by 0 there, so it gives no slope. The tangent line is vertical, and a vertical line has no slope.

Example 1: A slope at one value of t

Find the slope of the curve \(x = t^2 + 1\), \(y = t^3 - t\) at \(t = 2\).

  1. Find the two rates by differentiating \(x\) and \(y\) with respect to \(t\).

    \[\dfrac{dx}{dt} = 2t,\qquad \dfrac{dy}{dt} = 3t^2 - 1\]

  2. Substitute \(t = 2\) in both rates. Both are positive, so the point moves to the right and up. The triangle is drawn at half size to fit.

    \[\cg{\dfrac{dx}{dt} = 4},\qquad \cg{\dfrac{dy}{dt} = 11}\]

  3. Divide the vertical rate by the horizontal rate. The tangent line rises 11 for every 4 across.

    \[\cy{\dfrac{dy}{dx}} = \dfrac{11}{4}\]

Example 2: A point that moves down

Find the slope of the curve \(x = t^2 + 2t\), \(y = t^3 - 6t\) at \(t = 1\).

  1. Differentiate \(x\) and \(y\) with respect to \(t\).

    \[\dfrac{dx}{dt} = 2t + 2,\qquad \dfrac{dy}{dt} = 3t^2 - 6\]

  2. Substitute \(t = 1\). The point moves 4 to the right and 3 down, so the vertical leg is red.

    \[\cg{\dfrac{dx}{dt} = 4},\qquad \cr{\dfrac{dy}{dt} = -3}\]

  3. A tempting wrong move: The slope is how fast \(y\) changes, so the slope is \(\frac{dy}{dt} = -3\).

    \[\dfrac{dy}{dx}\ \overset{?}{=}\ \cr{-3}\]

    The rate \(-3\) is how fast \(y\) changes per unit of \(t\), not per unit of \(x\). The red line falls 3 for every 1 across, but the point falls only 3 for every 4 across.

  4. Divide the vertical rate by the horizontal rate. The slope is negative, because the point moves down as it moves to the right.

    \[\cy{\dfrac{dy}{dx}} = \dfrac{-3}{4} = -\dfrac34\]

Practice

A point moves with \(x = 2 + \sin t\) and \(y = \cos 2t\). Find the slope of its path at \(t = \frac{\pi}{6}\).

The triangle at the orange point has a horizontal leg \(\frac{dx}{dt}\) and a vertical leg \(\frac{dy}{dt}\). Find both at \(t = \frac{\pi}{6}\). Then divide the vertical leg by the horizontal leg.

Solution

  1. Differentiate both functions with respect to \(t\). The second one needs the chain rule.

    \[\dfrac{dx}{dt} = \cos t,\qquad \dfrac{dy}{dt} = -2\sin 2t\]

  2. Substitute \(t = \frac{\pi}{6}\), so \(2t = \frac{\pi}{3}\).

    \[\cg{\dfrac{dx}{dt} = \dfrac{\sqrt3}{2}},\qquad \cr{\dfrac{dy}{dt} = -\sqrt3}\]

  3. Divide. The slope is negative, because the point moves to the right and down.

    \[\cy{\dfrac{dy}{dx}} = \dfrac{-\sqrt3}{\sqrt3/2} = -2\]

Questions

Questions 1 and 2 use the curve \(x = t^2\), \(y = t^3 - 4t\). The orange point is where \(t = 1\), and the white point is where \(t = -1\). Each point has its triangle of the two rates, drawn to the scale of the grid.

  1. What is the slope of the curve at the orange point, where t = 1?

    • \(-\frac12\)
    • \(-2\)
    • \(\frac12\)
    Answer

    \(-\frac12\)

  2. What is the slope of the curve at the white point, where t = −1?

    • \(\frac12\)
    • \(-\frac12\)
    • 2
    Answer

    \(\frac12\)

  3. A point moves with \(x = e^{2t}\) and \(y = e^{3t}\). What is the slope of its path at \(t = \ln 2\)?

    • 3
    • \(\frac32\)
    • \(\frac13\)
    Answer

    3

Summary

You find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\), substitute the value of \(t\), and divide the second by the first.

This works because the point moves \(\frac{dy}{dt}\) up for every \(\frac{dx}{dt}\) across, and the chain rule gives the same ratio.

You use it when a curve is given by \(x = f(t)\) and \(y = g(t)\), at a value of \(t\) where \(\frac{dx}{dt}\) is not 0.

This is one of four skills in topic 9.1 of Which & Why: AP Calculus.