Wemazi

Topic 2.5

Use a slope to determine an unknown

Builds on evaluating the slope of a power function at a point. 6 minutes.

Tuning a coefficient

The curve is \(y = ax^2\), where the coefficient \(a\) is a number that is not known yet. The orange point is at \(x = 2\), and the orange line is the tangent there.

The lower graph shows the slope of the curve at each \(x\), and its orange bar is the slope at \(x = 2\).

The tangent must have slope 10, which the faint dashed line through the point shows. On the lower graph, the dashed line at height 10 marks the same target.

Slide \(a\). At \(a = 1\) the slope is 4, and at \(a = 2\) it is 8, so the slope is \(4a\).

The tangent lies on the dashed line when \(a = 2.5\), because \(4 \cdot 2.5 = 10\).

One slope condition, one equation

The curve is \(y = ax^2\) with an unknown coefficient \(a\). The point is at \(x = p\), and the dashed line shows the slope \(m\) that the tangent must have.

The slope at \(p\) is \(2ap\), which is \(a\) times the slope of \(x^2\) at \(p\). Slide \(a\), and the tangent turns.

Only one value of \(a\) puts the tangent on the dashed line. If you move \(p\) or \(m\), that value changes, but there is still only one. A slope condition is one equation, so it fixes one unknown.

Notation

  1. Differentiate with \(a\) as a constant factor. The unknown stays in the formula.

    \[y = ax^2 \;\Rightarrow\; y' = 2a\,x\]

  2. The condition "the slope at \(x = 2\) is 10" means \(y'(2) = 10\). Substitute \(x = 2\) into \(y'\). The two dashed lines show the target.

    \[y'(2) = 2a \cdot 2 = \cy{4a} = 10\]

  3. This is an equation in \(a\). Solve it, and the tangent turns onto the dashed line.

    \[a = \frac{10}{4} = \frac52\]

  4. Test the answer. With \(a = \tfrac52\) the slope formula is \(y' = 5x\), and \(y'(2) = 10\).

  5. A point on the curve gives a different equation. "The curve passes through \((1, 3)\)" is a condition on \(y\), not on \(y'\). It gives \(a \cdot 1^2 = 3\), so \(a = 3\).

    \[y(1) = a \cdot 1^2 = 3\]

When it fails

The slope must depend on \(a\) at the chosen point. At \(x = 0\) the slope of \(ax^2\) is \(2a \cdot 0 = 0\) for every \(a\).

Move the point to 0 and slide \(a\). The tangent stays flat. A slope of 10 there gives the equation \(0 = 10\), which has no solution.

Example 1: a coefficient

The curve \(y = ax^3\) has slope 24 at \(x = 2\). Find \(a\).

  1. Differentiate with \(a\) in place. The exponent 3 comes down.

    \[y' = 3a\,x^2\]

  2. The two dashed lines show that the slope at \(x = 2\) must be 24. Substitute \(x = 2\).

    \[3a \cdot 2^2 = \cy{12a} = 24\]

  3. Divide both sides by 12.

    \[a = 2\]

  4. With \(a = 2\) the slope formula is \(6x^2\), and it gives 24 at \(x = 2\). The tangent lies on the dashed line.

Example 2: the place where the slope has a given value

Find the points on \(y = \dfrac{2}{x}\) where the slope is \(-\tfrac12\).

  1. Write the curve as a power of \(x\). Then differentiate.

    \[\begin{aligned} y = 2x^{-1} \;\Rightarrow\; y' &= -2x^{-2} \\ &= -\frac{2}{x^2} \end{aligned}\]

  2. Set the slope equal to \(-\tfrac12\), and solve for \(x\). On the lower graph, the dashed line at height \(-\tfrac12\) meets the slope curve where this holds.

    \[-\frac{2}{x^2} = -\frac12 \;\Rightarrow\; x^2 = 4\]

  3. A tempting wrong move: Take the square root: \(x = 2\).

    \[x = 2\]

    The curve has two branches, and both fall. The lower dashed line meets the slope curve twice, so a second point with this slope is at \(x = -2\).

  4. The equation \(x^2 = 4\) has two solutions, 2 and \(-2\), because \((-2)^2 = 4\) too. Substitute each into \(y = \tfrac{2}{x}\) to find the points.

    \[(2, 1) \text{ and } (-2, -1)\]

Practice

The curve \(y = a\sqrt{x}\) has slope 3 at \(x = 4\). Find \(a\).

The slope of \(a\sqrt{x}\) is \(a\) times the slope of \(\sqrt{x}\). Slide \(a\) until the orange tangent lies on the dashed line.

Solution

  1. Write the root as a power, and differentiate with \(a\) in place.

    \[y = a\,x^{1/2} \;\Rightarrow\; y' = \frac{a}{2}\,x^{-1/2}\]

  2. Substitute \(x = 4\). The slope must be 3.

    \[\frac{a}{2} \cdot \frac{1}{2} = \frac{a}{4} = 3\]

  3. Multiply both sides by 4.

    \[a = 12\]

Questions

Questions 1 and 2 use the graph of a curve \(y = ax^3\) and its slope. Question 3 needs a calculation.

  1. The curve is \(y = ax^3\). At \(x = 1\) the orange bar reaches 6, so the tangent has slope 6. What is \(a\)?

    • 2
    • 6
    • 18
    • 3
    Answer

    2

  2. The dashed line on the lower graph is at height 24. At which values of \(x\) does the curve have slope 24?

    • Only \(x = 2\)
    • \(x = 2\) and \(x = -2\)
    • \(x = 4\) and \(x = -4\)
    Answer

    \(x = 2\) and \(x = -2\)

  3. The curve \(y = \dfrac{a}{x}\) has slope \(-3\) at \(x = 2\). Find \(a\).

    • 12
    • −12
    • 6
    • \(\tfrac34\)
    Answer

    12

Summary

You find an unknown coefficient by keeping it in the derivative, substituting the \(x\) of the point, and solving for it.

This works because the slope at a point is the derivative at that \(x\), so each slope condition becomes one equation.

One condition fixes one unknown, provided the slope depends on that unknown at the chosen point.

This is one of four skills in topic 2.5 of Which & Why: AP Calculus.