Wemazi

Topic 2.8

Apply a product derivative

Builds on using a product rule with data. 8 minutes.

A total that is a product

A fund holds pledges from donors. On day \(t\) there are \(N=300-5t\) donors, and the average pledge is \(A=40+2t\) dollars.

The rectangle has width \(N\) and height \(A\), so its area is the total pledged, \(T=NA\). At \(t=5\) the width is 275 and the height is 50, so \(T=13{,}750\) dollars.

Move the slider. The width shrinks, because 5 donors leave each day. The height grows, because the average pledge rises by 2 dollars each day.

The red strip along the right edge shows what the shrinking width removes in one day: 5 · 50 = 250 dollars. The green strip on top shows what the growing height adds: 275 · 2 = 550 dollars.

At day 5 the total changes by 550 − 250 = 300 dollars per day, so it grows although the number of donors falls.

The rate of a product of changing quantities

A total \(T\) is the product of a count \(N\) and an average \(A\), and both change with time.

When \(N\) changes at the rate \(N'\), a strip of height \(A\) appears on the right. When \(A\) changes at the rate \(A'\), a strip of width \(N\) appears on top.

Move the sliders. A strip is green when its rate is positive and red when it is negative, because then it removes area. For any rates, the rate of \(T\) is the sum of the two strips, with their signs.

Notation

  1. If \(T=NA\) and both \(N\) and \(A\) are differentiable, the rate of change of \(T\) is the sum of the two strips.

    \[T' = \cy{N'}\,A + N\,\cy{A'}\]

  2. The term \(N'A\) is the part of the change in \(T\) that comes from \(N\), and \(NA'\) is the part that comes from \(A\).

  3. The same rule gives the slope of the tangent to \(y=u(x)\,v(x)\) at \(x=a\).

    \[y'(a) = u'(a)\,v(a) + u(a)\,v'(a)\]

  4. Give the units. Here \(N'A\) is in donors per day times dollars per donor, which is dollars per day.

Example 1: a tangent line

Find the equation of the tangent line to \(y=(x+2)\cdot3^x\) at \(x=0\).

  1. The point of tangency has \(y=2\), because \((0+2)\cdot3^0=2\).

    \[y(0)=2\]

  2. Let \(u=x+2\) and \(v=3^x\). Then \(u'=1\) and \(v'=3^x\ln 3\).

    \[u'=1, \qquad v'=3^x\ln 3\]

  3. By the product rule, \(y'=u'v+uv'\).

    \[y' = 3^x + (x+2)\,3^x\ln 3\]

  4. Substitute \(x=0\). As a check, \(1+2\ln 3\approx3.20\), the height of the orange bar on the slope graph below.

    \[y'(0) = 1 + 2\ln 3\]

  5. The line passes through (0, 2) with this slope.

    \[y = 2 + (1+2\ln 3)\,x\]

Example 2: a rate with units

A fish farm has \(N=600\) fish, with the average mass \(m=1.5\) kg. The number grows by 40 fish per week. The average mass grows by 0.1 kg per week. How fast does the total mass \(B=Nm\) change?

  1. The total mass is \(B=Nm=900\) kg, the area of a rectangle with width \(N\) and height \(m\).

  2. A tempting wrong move: Multiply the two rates.

    \[\cr{B' = N'\cdot m' = 40\cdot0.1 = 4}\]

    Each rate is per week, so their product is per week per week, not kg per week. It is the small red dashed rectangle at the corner, not one of the two green strips of 60 that the rule adds.

  3. Use the product rule. The right strip is the new fish times the average mass. The top strip is the number of fish times the gain in average mass.

    \[\begin{aligned} B' &= 40\cdot1.5 + 600\cdot0.1 \\ &= 60+60 = 120 \end{aligned}\]

  4. The total mass grows by 120 kg per week, half from new fish and half from heavier fish.

Practice

A rectangle has length 12 cm, which grows at 3 cm per second. Its width is 5 cm, which shrinks at 2 cm per second. Find the rate of change of the area.

The growing length makes a green strip, and the shrinking width makes a red strip. Add them with their signs.

Solution

  1. The area is \(A=\ell w\), so the product rule gives its rate.

    \[A' = \ell' w + \ell\, w'\]

  2. Substitute the four numbers, with \(w'=-2\) because the width shrinks.

    \[A' = 3\cdot5 + 12\cdot(-2) = -9\]

  3. The area decreases by 9 cm² per second, because the red strip, 24, is larger than the green strip, 15.

Questions

Questions 1 and 2 use the donor fund of the opening, where \(N=300-5t\) and \(A=40+2t\). The strips show the change in one day. Question 3 needs a calculation.

  1. At \(t=5\) the red strip shows the dollars lost in one day as donors leave. How large is it?

    • 550 dollars
    • 300 dollars
    • 250 dollars
    Answer

    250 dollars

  2. At \(t=20\) the green strip and the red strip have the same area. What does this show about \(T\)?

    • \(T=0\)
    • \(T'=0\), so \(T\) is not changing at that moment
    • \(N\) and \(A\) are not changing
    Answer

    \(T'=0\), so \(T\) is not changing at that moment

  3. What is the slope of the tangent to \(y=(3x-2)\sin x\) at \(x=\frac{\pi}{2}\)?

    • 0
    • \(\frac{3\pi}{2}-2\)
    • 3
    Answer

    3

Summary

You find the rate of a product by adding two parts, and each part keeps one quantity and uses the rate of the other.

This works because the total changes when either quantity changes, and each change adds or removes one strip of area.

You use it for a tangent slope or a rate when a quantity is a product of two changing quantities.

This is one of four skills in topic 2.8 of Which & Why: AP Calculus.