Wemazi

Topic 9.1 · BC

Find the point and tangent line

Builds on calculating the slope of a parametric curve. 7 minutes.

A point and a slope from one moment

A point moves along the curve \(x = \sin 2t\), \(y = \sin t\), which has the shape of an 8. The orange point shows its position at time \(t\).

The orange line is the tangent line at that point. It touches the curve there and has the direction of the motion.

At \(t = \frac{\pi}{6}\), the position is \(\left(\frac{\sqrt3}{2}, \frac12\right)\). There \(\frac{dx}{dt} = 1\) and \(\frac{dy}{dt} = \frac{\sqrt3}{2}\), so the slope is \(\frac{\sqrt3}{2}\).

Move \(t\). The point and the line move together, because the point and the slope come from the same value of \(t\).

A tangent line has two parts

The orange point \(P\) is the position at a value \(t_0\) of the parameter, and the faint point \(Q\) is at another value.

A tangent line has two parts. The point \((x(t_0), y(t_0))\) shows where the line touches the curve, and the slope at \(t_0\) shows how steep it is.

Press the button. The red line passes through \(P\) but has the slope that belongs to \(Q\), so it crosses the curve instead of touching it. The point and the slope must come from the same value of \(t\).

Notation

  1. Choose the value of the parameter and call it \(t_0\). The orange point is the position at this value.

  2. Find the point by substituting \(t_0\) in \(x\) and in \(y\).

    \[(x_0,\,y_0) = \cy{\big(x(t_0),\,y(t_0)\big)}\]

  3. Find the slope by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\), both at \(t_0\).

    \[m = \cy{\dfrac{dy/dt}{dx/dt}}\Big|_{t=t_0}\]

  4. Write the line through the point with this slope. The line contains no \(t\), because \(x_0\), \(y_0\) and \(m\) are numbers.

    \[y - y_0 = m\,(x - x_0)\]

  5. The slope exists only if \(\frac{dx}{dt}\) is not 0 at \(t_0\).

When it fails

The moving point passes through the center twice, at \(t = 0\) and at \(t = \pi\), so one point on the curve belongs to two values of \(t\). Move the point to the center.

At \(t = 0\) the slope is \(\frac12\), and at \(t = \pi\) it is \(-\frac12\), so the center has two tangent lines. If a point belongs to several values of \(t\), write one line for each value.

Example 1: The tangent at t = 1

Find the tangent line to \(x = 3t^2\), \(y = 2t^3 - t\) at \(t = 1\).

  1. Find the point by substituting \(t = 1\) in \(x\) and in \(y\).

    \[(x_0,\,y_0) = \cy{(3,\,1)}\]

  2. Find the rates at \(t = 1\). Their ratio is the slope.

    \[\begin{aligned}\dfrac{dx}{dt} &= 6t = 6, \\ \dfrac{dy}{dt} &= 6t^2 - 1 = 5, \\ \cy{m} &\cy{{}= \dfrac56}\end{aligned}\]

  3. Write the line through \((3, 1)\) with slope \(\frac56\), and simplify.

    \[\begin{aligned}y - 1 &= \dfrac56(x - 3) \\ \Rightarrow\ y &= \dfrac56x - \dfrac32\end{aligned}\]

Example 2: A point given by coordinates

Find the tangent lines to \(x = t^3 - t\), \(y = t^2 + 1\) at the point \((0, 2)\).

  1. The point is given by coordinates and not by a value of \(t\), so first find the values of \(t\) that give this point.

  2. A tempting wrong move: The \(x\)-coordinate is 0 when \(t = 0\), so the point belongs to \(t = 0\).

    \[x = t^3 - t = 0\ \Rightarrow\ \cr{t = 0}\]

    At \(t = 0\), \(y\) is 1, so the curve is at \((0, 1)\) and not at \((0, 2)\). The red tangent line touches the curve below the orange point.

  3. Both coordinates must match. The equation \(t^3 - t = 0\) gives \(t = 0, 1, -1\), and \(t^2 + 1 = 2\) gives \(t = 1, -1\). Only \(t = 1\) and \(t = -1\) solve both, so the curve passes through \((0, 2)\) twice.

    \[t = 1\quad\text{and}\quad t = -1\]

  4. Find the slope at each value by dividing the rates.

    \[\begin{aligned}&\dfrac{dy}{dx} = \dfrac{2t}{3t^2 - 1}: \\ &\cy{m = 1}\ (t = 1), \\ &\cy{m = -1}\ (t = -1)\end{aligned}\]

  5. Write one line through \((0, 2)\) for each slope.

    \[y = x + 2,\qquad y = -x + 2\]

Practice

Find the tangent line to \(x = t^2 - 1\), \(y = \dfrac1t\) at \(t = 2\).

Move the slider to t = 2. The orange point is where the line touches the curve, and the slope must come from the same value of t.

Solution

  1. Find the point at \(t = 2\).

    \[(x_0,\,y_0) = \cy{\left(3,\,\dfrac12\right)}\]

  2. Find the rates at \(t = 2\). Then divide.

    \[\begin{aligned}\dfrac{dx}{dt} &= 2t = 4, \\ \dfrac{dy}{dt} &= -\dfrac1{t^2} = -\dfrac14, \\ \cy{m} &\cy{{}= -\dfrac1{16}}\end{aligned}\]

  3. Write the line and simplify.

    \[\begin{aligned}y - \dfrac12 &= -\dfrac1{16}(x - 3) \\ \Rightarrow\ y &= -\dfrac{x}{16} + \dfrac{11}{16}\end{aligned}\]

Questions

Questions 1 and 2 use the curve \(x = t - \sin t\), \(y = 1 - \cos t\), the path of a point on the rim of a rolling wheel. The orange point shows the position where \(t = \frac{\pi}{2}\). The white point marks the top of the first arch, where \(t = \pi\).

  1. Which line is tangent to the curve at the orange point, where \(t = \frac{\pi}{2}\)?

    • \(y = x - \frac{\pi}{2} + 2\)
    • \(y = x - \frac{\pi}{2} + 1\)
    • \(y = x\)
    Answer

    \(y = x - \frac{\pi}{2} + 2\)

  2. Which line is tangent to the curve at the white point, where \(t = \pi\)?

    • \(y = 2\)
    • \(x = \pi\)
    • \(y = 0\)
    Answer

    \(y = 2\)

  3. Find the tangent line to \(x = 2t\), \(y = t^2 + t\) at \(t = 1\).

    • \(y = \frac32x - 1\)
    • \(y = \frac23x + \frac23\)
    • \(y = \frac32x + \frac12\)
    Answer

    \(y = \frac32x - 1\)

Summary

You find the point by substituting the value of \(t\) in \(x\) and \(y\), and the slope by dividing \(\frac{dy}{dt}\) by \(\frac{dx}{dt}\) there.

The line through this point with this slope touches the curve, because the point and the slope describe the same moment of the motion.

If the point is given by coordinates, you first find every value of \(t\) that gives it and write one line for each.

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