Wemazi

Topic 9.1 · BC

Locate horizontal and vertical tangents

Builds on finding the point and tangent line. 8 minutes.

When one leg shrinks to nothing

A point moves around the ellipse \(x = 4\cos t\), \(y = 3\sin t\). The triangle at the orange point has a horizontal leg \(\frac{dx}{dt}\) and a vertical leg \(\frac{dy}{dt}\). Each leg is green when it points up or to the right, and red when it points down or to the left.

Drag the point to the top, where \(t = \frac{\pi}{2}\). There \(\frac{dy}{dt} = 0\), so the vertical leg shrinks to 0, the point moves straight across and the tangent line is horizontal.

Drag the point to the right end, where \(t = 0\). There \(\frac{dx}{dt} = 0\), so the horizontal leg shrinks to 0, the point moves straight up and the tangent line is vertical.

At each place the other leg is not 0, so it gives the direction of the motion.

A zero in one rate, not in both

Below the ellipse, two graphs show \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) against \(t\), green where the rate is positive and red where it is negative. The orange bar marks the current value of \(t\).

Move \(t\). Where the \(\frac{dy}{dt}\) graph crosses the axis, the vertical leg is 0 and the tangent is horizontal. Where the \(\frac{dx}{dt}\) graph crosses the axis, the horizontal leg is 0 and the tangent is vertical.

A zero in one rate gives a direction only while the other rate is not 0. If both rates are 0, the triangle shrinks to a point and the motion has no direction.

Notation

  1. For a horizontal tangent, solve \(\frac{dy}{dt} = 0\). Then check that \(\frac{dx}{dt}\) is not 0 at each solution.

    \[\cy{\dfrac{dy}{dt} = 0}\quad\text{and}\quad \dfrac{dx}{dt} \neq 0\]

  2. For a vertical tangent, solve \(\frac{dx}{dt} = 0\). Then check that \(\frac{dy}{dt}\) is not 0 at each solution.

    \[\cy{\dfrac{dx}{dt} = 0}\quad\text{and}\quad \dfrac{dy}{dt} \neq 0\]

  3. Both tests come from the slope \(\frac{dy/dt}{dx/dt}\). If only the top is 0, the slope is 0. If only the bottom is 0, the slope is undefined and the tangent is vertical.

  4. A zero of \(\frac{dx}{dt}\) alone gives a vertical tangent, not a horizontal one.

  5. To find the point, substitute each value of \(t\) into \(x\) and \(y\).

When it fails

On the curve \(x = 4t^2\), \(y = 3t^3\), the rates are \(8t\) and \(9t^2\), so both are 0 at \(t = 0\). Switch to this curve and move the point to \(t = 0\). The triangle shrinks to a point, because the point stops for an instant.

The test needs one rate that is not 0. Here neither rate is, so the test does not decide. To find the direction, find what the slope approaches near that value of \(t\).

Example 1: Several horizontal tangents

Find the points on \(x = t^2 - 4t + 3\), \(y = t^4 - 2t^2\) where the tangent is horizontal or vertical.

  1. Find both rates and factor the second one.

    \[\begin{aligned} \dfrac{dx}{dt} &= 2t - 4 \\ \dfrac{dy}{dt} &= 4t^3 - 4t = 4t(t - 1)(t + 1) \end{aligned}\]

  2. For horizontal tangents, solve \(\frac{dy}{dt} = 0\), which gives \(t = 0, 1\) and \(-1\). At these values \(\frac{dx}{dt}\) is \(-4\), \(-2\) and \(-6\), so it is not 0.

    \[\begin{aligned} t = 0&:\ (3,\,0) \\ t = 1&:\ (0,\,-1) \\ t = -1&:\ (8,\,-1) \end{aligned}\]

  3. For vertical tangents, solve \(\frac{dx}{dt} = 0\), which gives \(t = 2\). There \(\frac{dy}{dt} = 24\), which is not 0, so the tangent is vertical.

    \[t = 2:\ (-1,\,8)\]

Example 2: A point where the motion stops

Find the horizontal and vertical tangents of \(x = t^3 - 3t^2\), \(y = t^2 - 4t\).

  1. Find both rates and factor them.

    \[\begin{aligned} \dfrac{dx}{dt} &= 3t(t - 2) \\ \dfrac{dy}{dt} &= 2(t - 2) \end{aligned}\]

  2. A tempting wrong move: The rate \(\frac{dy}{dt}\) is 0 at \(t = 2\), so the tangent at \((-4, -4)\) is horizontal.

    \[\dfrac{dy}{dt} = 0\ \Rightarrow\ \cr{\text{horizontal}}\]

    The rate \(\frac{dx}{dt}\) is also 0 at \(t = 2\), so the triangle has no sides and the test does not decide. Near this point the curve rises 1 for every 3 across, so the red horizontal line is not tangent.

  3. For \(t \neq 2\), cancel the factor \(t - 2\) in the ratio. As \(t\) approaches 2, the slope approaches \(\frac13\), so the tangent at \((-4, -4)\) has slope \(\frac13\).

    \[\dfrac{dy}{dx} = \dfrac{2(t - 2)}{3t(t - 2)} = \dfrac{2}{3t}\ \to\ \cy{\dfrac13}\]

  4. The rate \(\frac{dy}{dt}\) is 0 only at \(t = 2\), so there is no horizontal tangent. The rate \(\frac{dx}{dt}\) is 0 at \(t = 0\), where \(\frac{dy}{dt} = -4\), so the tangent at the origin is vertical.

    \[t = 0:\ (0,\,0)\]

Practice

Find the points on \(x = te^{-t}\), \(y = t^2 - 4t\), with \(t \ge 0\), where the tangent is horizontal or vertical.

Move t until a leg of the triangle shrinks to 0. Solve each rate for 0 separately. Then check the other rate at each solution.

Solution

  1. Find both rates.

    \[\begin{aligned} \dfrac{dx}{dt} &= (1 - t)e^{-t} \\ \dfrac{dy}{dt} &= 2t - 4 \end{aligned}\]

  2. The rate \(\frac{dy}{dt}\) is 0 at \(t = 2\). The rate \(\frac{dx}{dt}\) there is \(-e^{-2}\), so the tangent is horizontal.

    \[t = 2:\ \cy{\left(\dfrac{2}{e^2},\,-4\right)}\]

  3. The rate \(\frac{dx}{dt}\) is 0 only at \(t = 1\), because \(e^{-t}\) is never 0. The rate \(\frac{dy}{dt}\) there is \(-2\), so the tangent is vertical.

    \[t = 1:\ \cy{\left(\dfrac{1}{e},\,-3\right)}\]

Questions

Questions 1 and 2 use the curve \(x = t^3 - 3t\), \(y = t^2\). The orange point is at the bottom of the curve, 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. At the orange point at the bottom of the curve, which rate is 0?

    • \(\frac{dy}{dt}\)
    • \(\frac{dx}{dt}\)
    • Both rates
    Answer

    \(\frac{dy}{dt}\)

  2. What is the tangent line at the white point, where t = −1, the point (2, 1)?

    • A vertical line
    • A horizontal line
    • There is no tangent line
    Answer

    A vertical line

  3. For \(x = t^2 - 6t\), \(y = t^3 - 3t^2 - 9t\), which value of \(t\) gives a horizontal tangent?

    • \(t = -1\) only
    • \(t = 3\) only
    • \(t = -1\) and \(t = 3\)
    Answer

    \(t = -1\) only

Summary

You find a horizontal tangent by solving \(\frac{dy}{dt} = 0\) and a vertical tangent by solving \(\frac{dx}{dt} = 0\).

This works because the slope is \(\frac{dy/dt}{dx/dt}\), which is 0 when only its top is 0 and undefined when only its bottom is 0.

At each solution you check that the other rate is not 0, because when both rates are 0 the test does not decide.

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