Use a quotient derivative at a point
A typing speed that levels off
A student practices typing. After \(t\) weeks the speed is \(W(t) = \dfrac{60t}{t+3}\) words per minute, for \(t \ge 0\). The graph rises and then flattens, and the faint dashed line marks 60 words per minute, which the speed does not reach.
Set \(t\) to 3. The speed is \(W(3) = 180/6 = 30\) words per minute.
The quotient rule gives \(W'(t) = \dfrac{180}{(t+3)^2}\). At \(t = 3\) this is \(180/36 = 5\), which is the slope of the orange tangent.
The slope has the unit words per minute, per week, so at week 3 the speed grows by about 5 words per minute each week.
Set \(t\) to 9. The slope is \(180/144 = 1.25\), so the curve is flatter.
Any quotient curve
The curve is a quotient. The dashed vertical line marks where the bottom is 0, and the curve has no point on that line.
Drag the point along either branch. The orange line is the tangent at the point, and its slope is the derivative at that input.
The input must be in the domain of the original function, so the point is never on the dashed line.
Notation
Choose the input \(a\). Check that \(a\) is in the domain of the original function, which means that the bottom is not 0 there. In a model, the input must also make sense, for example \(t \ge 0\).
Find \(f'\) with the quotient rule. Then substitute \(a\), and the result is the slope of the tangent at the point \((a, f(a))\).
\[m = \cy{f'(a)}\]The tangent line passes through the point with the slope \(m\).
\[y - f(a) = \cy{m}\,(x - a)\]The normal line is the line perpendicular to the tangent at the point. Its slope is the negative reciprocal of \(m\), if \(m\) is not 0.
\[-\frac{1}{m}\]In a model, the unit of the slope is the unit of the output for each unit of the input.
When it fails
At the dashed line, the bottom is 0. The original function has no value there, so the curve has no point and no tangent.
The quotient rule divides by the square of that bottom, so there is no derivative there either.
Example 1: tangent and normal lines
Find the tangent line and the normal line to \(y = \dfrac{x+3}{x-1}\) at \(x = 3\).
The bottom at \(x = 3\) is \(3 - 1 = 2\), not 0, so 3 is in the domain. The point has \(y = 6/2 = 3\).
\[(3,\ 3)\]The top of the rule is \((x-1) - (x+3) = -4\).
\[\begin{aligned}y' &= \frac{(1)(x-1) - (x+3)(1)}{(x-1)^2} \\ &= \frac{-4}{(x-1)^2}\end{aligned}\]At \(x = 3\) the bottom of the rule is \((3-1)^2 = 4\), so the slope is \(-4/4\).
\[m = \cy{-1}\]Write the tangent line through \((3, 3)\) with the slope \(-1\).
\[y - 3 = -(x - 3), \quad y = -x + 6\]The normal slope is the negative reciprocal of \(-1\), which is 1. The normal line passes through the same point.
\[y - 3 = 1\,(x - 3), \quad y = x\]
Example 2: where does the slope equal 1/4?
Find the points on \(y = \dfrac{x}{x+1}\) where the tangent has slope \(\tfrac14\). The function is defined for \(x \ne -1\).
The quotient rule with the top \(x\) and the bottom \(x+1\) gives the slope, which the lower graph shows at each \(x\).
\[y' = \frac{(x+1) - x}{(x+1)^2} = \frac{1}{(x+1)^2}\]Set the slope equal to \(\tfrac14\). On the lower graph, the answers are where the slope curve crosses the dashed level at \(\tfrac14\).
\[\begin{aligned}\frac{1}{(x+1)^2} &= \frac14 \\ (x+1)^2 &= 4\end{aligned}\]A tempting wrong move: Take the square root. Then \(x + 1 = 2\), so \(x = 1\) is the only point.
\[x + 1 = 2\]A square has two roots, 2 and \(-2\). The slope curve crosses the level a second time, at \(x = -3\), on the other branch. The red text marks the point that this step loses.
Use both roots, \(x + 1 = 2\) or \(x + 1 = -2\). Neither answer is \(-1\), so both are in the domain.
\[x = 1 \quad\text{or}\quad x = -3\]Then \(y(1) = \tfrac12\) and \(y(-3) = \tfrac{-3}{-2} = \tfrac32\), so the points are \((1, \tfrac12)\) and \((-3, \tfrac32)\).
Practice
The concentration of a medicine in the blood is \(P(t) = \dfrac{12t}{t^2+4}\) milligrams per liter, \(t\) hours after a dose, for \(t \ge 0\). Find \(P'(4)\) and say what it means.
The curve rises first and then falls. At \(t = 4\) it is past the peak, so the slope must be negative.
Solution
The top \(12t\) has the derivative 12, and the bottom \(t^2+4\) has the derivative \(2t\). The bottom is never 0, so \(P'\) exists for every \(t\).
\[\begin{aligned}P'(t) &= \frac{12(t^2+4) - 12t\cdot 2t}{(t^2+4)^2} \\ &= \frac{12(4 - t^2)}{(t^2+4)^2}\end{aligned}\]At \(t = 4\) the top is \(12(4 - 16) = -144\), and the bottom is \(20^2 = 400\).
\[P'(4) = \frac{-144}{400} = -\frac{9}{25}\]The unit is milligrams per liter per hour, and \(-\tfrac{9}{25} = -0.36\). At four hours, the concentration falls by about 0.36 milligrams per liter each hour.
Questions
The figure shows \(f(x) = \dfrac{x-2}{x+1}\). The marked point moves to the input that each question names. Questions 1 and 2 use the graph. Question 3 needs a calculation.
The curve crosses the x-axis at the marked point \(x = 2\). What is the slope of the tangent there?
- \(\tfrac13\)
- \(-\tfrac13\)
- 1
- 3
Answer
\(\tfrac13\)
The marked point is now at \(x = -1\). What does the figure show there?
- No tangent line, because the curve has no point at \(x = -1\)
- A vertical tangent line
- A tangent line with slope 3
Answer
No tangent line, because the curve has no point at \(x = -1\)
Find the slope of the normal line to \(y = \dfrac{3x}{x+2}\) at \(x = 1\).
- \(-\tfrac32\)
- \(\tfrac23\)
- \(-\tfrac23\)
- \(\tfrac32\)
Answer
\(-\tfrac32\)
Summary
You check that the input is in the domain, find the derivative with the quotient rule, and substitute the input to find the slope.
That value is the slope of the tangent, so it gives the tangent line, the normal line, and a rate with its unit.
You use it when a question asks for the slope or the rate of a ratio at one input.
This is one of four skills in topic 2.9 of Which & Why: AP Calculus.