Wemazi

Topic 3.2

Find the tangent at a point

Builds on solving for the derivative. 7 minutes.

A point on the curve, and a point off it

The curve is \(x^3 + y^3 = 9\). The point \((1, 2)\) is on it, because \(1 + 8 = 9\), but \((2, 2)\) is not, because \(8 + 8 = 16\).

The slope formula is \(\tfrac{dy}{dx} = -\tfrac{x^2}{y^2}\). At \((1, 2)\) it gives \(-\tfrac14\), and the orange tangent passes through the point with that slope. Drag the point along the curve, and the tangent changes with it.

The formula also gives a number at \((2, 2)\), which is \(-1\). Show the line through \((2, 2)\) with that slope. The red line does not touch the curve, so it is not a tangent.

A tangent line needs a point on the curve and the slope at that point.

Any point on any curve

Remove the numbers. The curve is an equation in \(x\) and \(y\), and the orange point is \((a, b)\).

Drag the point. At each position, the tangent has the slope \(m = \tfrac{dy}{dx}\) at that point.

Two facts fix the tangent line: a point that satisfies the equation of the curve, and the slope there.

The slope formula uses both coordinates, so the point decides the slope.

Notation

  1. Substitute \(x = a\) and \(y = b\). If the equation holds, \((a, b)\) is on the curve.

  2. Find \(\tfrac{dy}{dx}\) by implicit differentiation. It is a formula in \(x\) and \(y\).

  3. Put in both coordinates of the point. The result is a number, the slope \(m\) at the point.

    \[m = \cy{\left.\frac{dy}{dx}\right|_{(a, b)}}\]

  4. The tangent line passes through \((a, b)\) with slope \(m\), so use the point-slope form.

    \[y - b = m\,(x - a)\]

    This needs \(\tfrac{dy}{dx}\) to exist, so its denominator must not be 0 at the point.

When it fails

Move the point to \((2, 2)\). The formula still gives a number, \(-1\), but the line with that slope does not touch the curve.

The formula gives the slope of the curve only at points of the curve, and a point off the curve has no tangent. Check the point first.

Example 1: An ellipse

Find the tangent line to \(2x^2 + y^2 = 27\) at \((3, 3)\).

  1. Check the point. It satisfies the equation, so it is on the curve.

    \[2(3)^2 + 3^2 = 18 + 9 = 27\]

  2. Differentiate both sides, and solve for the slope.

    \[\begin{aligned} 4x + 2y\,\frac{dy}{dx} &= 0 \\ \frac{dy}{dx} &= -\frac{2x}{y} \end{aligned}\]

  3. Put in \(x = 3\) and \(y = 3\).

    \[m = -\frac{2(3)}{3} = \cy{-2}\]

  4. Write the line through \((3, 3)\) with slope \(-2\).

    \[\begin{aligned} y - 3 &= -2(x - 3) \\ y &= -2x + 9 \end{aligned}\]

Example 2: A product term

Find the tangent line to \(x^3 + xy + y^3 = 11\) at \((2, 1)\).

  1. Check the point. It satisfies the equation, so it is on the curve.

    \[2^3 + 2(1) + 1^3 = 11\]

  2. Differentiate both sides. The product \(xy\) gives \(y + x\,\tfrac{dy}{dx}\), and \(y^3\) needs the chain rule.

    \[\begin{aligned} 3x^2 + y &+ x\,\frac{dy}{dx} \\ &+ 3y^2\,\frac{dy}{dx} = 0 \end{aligned}\]

  3. Collect the terms with \(\tfrac{dy}{dx}\) on one side, factor, and divide to solve for the slope.

    \[\frac{dy}{dx} = -\frac{3x^2 + y}{x + 3y^2}\]

  4. Put in \(x = 2\) and \(y = 1\).

    \[m = -\frac{12 + 1}{2 + 3} = \cy{-\tfrac{13}{5}}\]

  5. A tempting wrong move: Write the line as \(y = mx + b\), and use the \(y\)-value of the point for \(b\).

    \[y = -\tfrac{13}{5}\,x + 1\]

    The number 1 is the \(y\)-value of the point, not the \(y\)-intercept. The red line has the correct slope but misses the orange point, because at \(x = 2\) it gives \(y = -4.2\), not 1.

  6. Use the point-slope form, which puts the point on the line.

    \[\begin{aligned} y - 1 &= -\tfrac{13}{5}(x - 2) \\ y &= -\tfrac{13}{5}\,x + \tfrac{31}{5} \end{aligned}\]

Practice

Show that \((1, 2)\) is on the curve \(3x^2 - 2xy + y^2 = 3\). Then find the tangent line there.

Check the point first. Then drag the orange point to \((1, 2)\): the tangent falls at a 45-degree angle.

Solution

  1. Put in the point. The equation holds, so the point is on the curve.

    \[\begin{aligned} 3(1)^2 - 2(1)(2) + 2^2 &= 3 - 4 + 4 \\ &= 3 \end{aligned}\]

  2. Differentiate both sides. The product \(xy\) gives two terms.

    \[\begin{aligned} 6x - 2y &- 2x\,\frac{dy}{dx} \\ &+ 2y\,\frac{dy}{dx} = 0 \end{aligned}\]

  3. Collect the terms with \(\tfrac{dy}{dx}\) on one side, factor, and divide to solve for the slope.

    \[\frac{dy}{dx} = \frac{y - 3x}{y - x}\]

  4. At \((1, 2)\), the slope is \(\tfrac{2 - 3}{2 - 1} = -1\). Use the point-slope form.

    \[\begin{aligned} y - 2 &= -(x - 1) \\ y &= -x + 3 \end{aligned}\]

Questions

Questions 1 and 2 use the curve \(y^2 = x^3 + 1\), its tangent at \((2, 3)\), and a hollow dot at \((2, 2)\).

  1. What is the equation of the tangent line at \((2, 3)\)?

    • \(y = 2x - 1\)
    • \(y = 2x + 3\)
    • \(y = 2x + 7\)
    Answer

    \(y = 2x - 1\)

  2. A student asks for the tangent line at the hollow dot \((2, 2)\). What is the correct response?

    • There is no tangent, because \((2, 2)\) is not on the curve.
    • Put \((2, 2)\) in the slope formula and write the line.
    • Use the tangent at \((2, 3)\), because it has the same \(x\).
    Answer

    There is no tangent, because \((2, 2)\) is not on the curve.

  3. The point \((1, 2)\) is on \(x^2y + xy^2 = 6\). Find the slope of the tangent there.

    • \(-\tfrac85\)
    • \(-8\)
    • \(-2\)
    • \(\tfrac85\)
    Answer

    \(-\tfrac85\)

Summary

You check that the point is on the curve, put its coordinates in \(\tfrac{dy}{dx}\), and write the line through the point with that slope.

This works because \(\tfrac{dy}{dx}\) is the slope of the curve at points of the curve, and a point with a slope fixes a line.

You use the point-slope form for the line, and you use the formula only at points that satisfy the equation.

This is one of four skills in topic 3.2 of Which & Why: AP Calculus.