AP Calculus AB and BC: Chapter 2 – Differentiation :2.6 -The Tangent Lines and the Normal Lines Study Notes

The slope of the tangent line to the graph of f at the point P(a, f (a)) is the number

\(m=\underset{h\rightarrow 0}{lim}\frac{f\left ( a+h \right )-f\left ( a \right )}{h}\)

if the limit exists.

Recall from the definition of the derivative that this limit is just \({f}’\left ( a \right ).\)

In point slope form the equation of the tangent line is

\(y-f\left ( a \right )={f}’\left ( a \right )\left ( x-a \right )\)

The normal line to the graph of f at the point P(a, f (a)) is the line that passes through P and is perpendicular to the tangent line to the graph of f at P .

Example 1 

  • Write the equations of the tangent line and normal line to the graph of \(y=x-\frac{x^{2}}{10}\) at the point \(\left ( 4,\frac{12}{5} \right ).\)
▶️Answer/Explanation

Solution

\({y}’=1-\frac{2x}{10}=1-\frac{x}{5}\)

At the point \(\left ( 4,\frac{12}{5} \right )\), the slope is \({y}’|_{x=4}=1-\frac{4}{5}=\frac{1}{5}\) and the equation of tangent line at this point is \(y-\frac{12}{5}=\frac{1}{5}\left ( x-4 \right )\) or \(y=\frac{1}{5}x+\frac{8}{5}.\)

At the point \(\left ( 4,\frac{12}{5} \right )\), the slope e of the normal line is negative reciprocal of 1/5, or -5.

So the equation of normal line at this point is \(y-\frac{12}{5}=-5\left ( x-4 \right )\) or \(y=-5x+22.4.\)

Exercises – Tangent Lines and Normal Lines

Multiple Choice Questions

1. The equation of the line tangent to the graph of \(y=x\sqrt{3+x^{2}}\) at the point (1,2) is

(A) \(y=\frac{3}{2}x-\frac{1}{2}\)               (B) \(y=2x+\frac{1}{2}\)                (C) \(y=\frac{5}{2}x-\frac{1}{2}\)               (D) \(y=\frac{5}{2}x+\frac{1}{2}\)

▶️Answer/Explanation

Ans:

1. C

2. Which of the following is an equation of the line tangent to the graph of \(f\left ( x \right )=x^{2}-x\) at the point where \({f}’\left ( x \right )=3\) ?

(A) \(y=3x-2\)

(B) \(y=3x+2\)

(C) \(y=3x-4\)

(D) \(y=3x+4\)

▶️Answer/Explanation

Ans:

2. C

3. A curve has slope \(2x+x^{-2}\) at each point (x,u) on the curve. Which of the following is an equation for this curve if it passes through the point (1,3) ? 

(A) \(y=2x^{2}+\frac{1}{x}\)

(B) \(y=x^{2}-\frac{1}{x}+3\)

(C) \(y=x^{2}+\frac{1}{x}+1\)

(D) \(y=x^{2}-\frac{2}{x^{2}}+4\)

▶️Answer/Explanation

Ans:

3. B

4. An equation of the line normal to the graph of y = tan x, at the point \(\left ( \frac{\pi }{6},\frac{1}{\sqrt{3}} \right )\) is

(A) \(y-\frac{1}{\sqrt{3}}=-\frac{1}{4}\left ( x-\frac{\pi }{6} \right )\)

(B) \(y-\frac{1}{\sqrt{3}}=\frac{1}{4}\left ( x-\frac{\pi }{6} \right )\)

(C) \(y-\frac{1}{\sqrt{3}}=-\frac{3}{4}\left ( x-\frac{\pi }{6} \right )\)

(D) \(y-\frac{1}{\sqrt{3}}=\frac{3}{4}\left ( x-\frac{\pi }{6} \right )\)

▶️Answer/Explanation

Ans:

4. C

5. If \(2x+3y=4\) is an equation of the line normal to the graph of f at the point \(\left ( -1,2 \right )\), then \({f}’\left ( -1 \right )=\)

(A) \(-\frac{2}{3}\)               (B) \(\frac{1}{\sqrt{2}}\)                (C) \(\sqrt{2}\)               (D) \(\frac{3}{2}\)

▶️Answer/Explanation

Ans:

5. D

6. If \(2x-y=k\) is an equation of the line normal to the graph of \(f\left ( x \right )=x^{4}-x\), then k =

(A) \(\frac{23}{16}\)               (B) \(\frac{13}{18}\)                (C) \(\frac{15}{16}\)               (D) \(\frac{9}{8}\)

▶️Answer/Explanation

Ans:

6. A

Free Response Questions

7. Line \(\imath \) is tangent to the graph of \(y=x-\frac{x^{2}}{120}\) at the point P and intersects x-axis at ( 15,0) − as shown in the figure above.

(a) Find the x-coordinates of point P .

(b) Write an equation for line \(\imath \) .

(c) If the line of symmetry for the curve \(y=x-\frac{x^{2}}{120}\) intersects line \(\imath \) at point R , what is the length of \(\bar{QR}\)?

▶️Answer/Explanation

Ans:

7. (a) 30

7. (b) \(y=\frac{1}{2}x+\frac{15}{2}\)

7. (c) 7.5

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