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8.3: Tangent Lines and Rates of Change

Difficulty Level: At Grade Created by: CK-12

Learning objectives

• Demonstrate an understanding of the rate of change of a function.
• Demonstrate an understanding of the slope of the tangent line to the graph.
• Demonstrate an understanding of the instantaneous rate of change.

Introduction

Calculus is a branch of mathematics that deals with rates of change of a quantity. For example, a car speeding down the street, the inflation of currency, the number of bacteria in a culture, and the AC voltage of an electric signal are all examples of quantities that change with time. In this section, we will study the rate of change of a quantity and how it is related to the tangent lines on a curve.

The Tangent Line

Back in Lesson 8.1, we briefly discussed how a secant line could become a tangent line to a point on a graph. Recall from algebra, if points P(x0,y0) and Q(x1,y1) are two different points on the curve y = f(x), then the slope of the secant line connecting the two points is given by

msec\begin{align*}m_{sec}\end{align*} =y1y0x1x0=f(x1)f(x0)x1x0\begin{align*}= \frac {y_1-y_0}{x_1-x_0} = \frac {f(x_1)-f(x_0)}{x_1-x_0}\end{align*} (1)

As discussed earlier, if we let the point x1 approach xo then Q will approach P along the graph f and thus the slope of the secant line will gradually approach the slope of the tangent line as x1 approaches x0. Therefore, (1) becomes

msec\begin{align*}m_{sec}\end{align*} =limx1x0f(x1)f(x0)x1x0\begin{align*}= \lim_{x_1 \to x_0} \frac {f(x_1)-f(x_0)}{x_1-x_0} \end{align*} (2)

To simplify our notation, if we let h = x1x0, then x1 = x0 + h and x1 → x0 becomes equivalent to h → 0. This means that (2) becomes

msec\begin{align*}m_{sec}\end{align*} =limh0f(x0+h)f(x0)h\begin{align*}= \lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*}

The Slope of a Tangent Line

If the point P(x0,y0) is on the curve f, then the tangent line at Phas a slope that is given bymtan=limh0f(x0+h)f(x0)h\begin{align*}m_{tan}=\lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*}

provided that the limit exist.

Recall that the equation of the tangent line through point (x0, y0) with slope m is the point-slope form of a line: yy0 = mtan(xx0).

Example 1: Find line tangent to the curve f(x)=x3 that passes through point P (2,8).

Solution: Since P(x0, y0) = (2, 8), using the slope of the tangent equation we have

mtan\begin{align*}m_{tan}\end{align*} =limh0f(x0+h)f(x0)h\begin{align*}= \lim_{h \to 0} \frac {f(x_0+h)-f(x_0)}{h} \end{align*}
and we get
mtan\begin{align*}m_{tan}\end{align*} =limh0f(2+h)f(2)h\begin{align*}= \lim_{h \to 0} \frac {f(2+h)-f(2)}{h} \end{align*}
=limh0(h3+6h2+12h+8)8h\begin{align*}= \lim_{h \to 0} \frac {(h^3+6h^2+12h+8)-8}{h} \end{align*}
=limh0h3+6h2+12hh\begin{align*}= \lim_{h \to 0} \frac {h^3+6h^2+12h}{h} \end{align*}
=limh0(h2+6h+12)\begin{align*}= \lim_{h \to 0} (h^2+6h+12) \end{align*}
=12\begin{align*}= 12\end{align*}

Thus the slope of the tangent line is 12. Using the point-slope formula above, we find that the equation of the tangent line is y -8 = 12 (x -2) or y = 12 x -16.

What we are interested in next is to be able to know the slope of the tangent line at any point on the curve f. Such a formula would be the same formula that we are using except we would replace the constant x0 by the variable x. This yields,

mtan\begin{align*}m_{tan}\end{align*} =limh0f(x+h)f(x)h\begin{align*}= \lim_{h \to 0} \frac {f(x+h)-f(x)}{h} \end{align*}
We denote this formula by,
f(x)\begin{align*}f'(x)\end{align*} =limh0f(x+h)f(x)h\begin{align*}= \lim_{h \to 0} \frac {f(x+h)-f(x)}{h} \end{align*}

where f' (x) reads "f prime of x" The next example illustrate its usefulness.

Example 2: If f(x) = x2 − 3,find f' (x) and use the result to find the slope of the tangent line at x = 2 and x = −1.

Solution: Since f(x)=limh0f(x+h)f(x)h\begin{align*}f'(x)= \lim_{h \to 0} \frac {f(x+h)-f(x)}{h} \end{align*}then

f(x)\begin{align*}f'(x)\end{align*} =limh0[(x+h)23][x23]h\begin{align*}= \lim_{h \to 0} \frac {\left [ {(x+h)^2-3} \right ]- \left [ {x^2-3} \right ]}{h} \end{align*}
=limh0x2+2xh+h23x2+3h\begin{align*}= \lim_{h \to 0}\frac {x^2+2xh+h^2-3-x^2+3}{h}\end{align*}
=limh02xh+h2h\begin{align*}= \lim_{h \to 0} \frac {2xh+h^2}{h}\end{align*}
=limh0(2x+h)\begin{align*}= \lim_{h \to 0} (2x+h)\end{align*}
=2x\begin{align*}= 2x\end{align*}

To find the slope,we simply substitute x = 2 into the result f' (x):

f'(x) = 2x
f'(2) =2(2)
= 4
and
f'(x) =2x
f'(-1) =2(-1)
= -2

Thus slope of the tangent line at x = 2 and x = −1 are 4 and −2 respectively.

Example 3: Find the slope of the tangent line to the curve y = 1/x that passes through the point (1, 1).

Solution: Using the slope of the tangent formula,

f(x)\begin{align*}f'(x)\end{align*} =limh0f(x+h)f(x)h\begin{align*}= \lim_{h \to 0} \frac {f(x+h)-f(x)}{h}\end{align*}
and substituting y=1x\begin{align*}y=\frac{1}{x}\end{align*}
f(x)\begin{align*}f'(x)\end{align*} \begin{align*}= \lim_{h \to 0} \frac {\left ( \frac{1}{x+h} \right )-\frac {1}{x}}{h}\end{align*}
\begin{align*}= \lim_{h \to 0} \frac {\frac {x-x-h}{x \left ({x+h} \right)}}{h}\end{align*}
\begin{align*}= \lim_{h \to 0} \frac {x-x-h}{hx \left ({x+h} \right)}\end{align*}
\begin{align*}= \lim_{h \to 0} \frac {-h}{hx \left ({x+h} \right)}\end{align*}
\begin{align*}= \lim_{h \to 0} \frac {-1}{x \left ({x+h} \right)}\end{align*}
\begin{align*}= \frac {-1}{x^2}\end{align*}
For x=1, the slope is
\begin{align*}f'(x)\end{align*} \begin{align*}= \frac {-1}{1}=-1\end{align*}
\begin{align*}= -1\end{align*}

Thus the slope of the tangent line at x = 1 for the curve y = 1/x is m = −1. To find the equation of the tangent line, we simply use the point-slope formula,

y-y0 =m(x-x0)
Where (x0, y0) = (1, 1).
y- 1 = -1(x-1)
y = -x+1+1
y = -x+2

So the equation of the tangent line is y=-x+2.

Average Rate of Change and Instantaneous Rate of Change

One of the two primary concepts of calculus involves calculating the rate of change of one quantity with respect to another. For example, speed is defined as the rate of displacement with respect to time. If a person travels 120 miles in 4 hours, his speed is 120/4=30 mi/hr. This speed is called the average speed or the average rate of change of distance with respect to time. Of course the person who travels 120 miles at a rate of 30 mi/hr for 4 hr probably does not do so continuously. Though he probably slowed down or sped up during the 4-hour period, it does suffice to say that he traveled for 4 hours at an average rate of 30 miles per hour. However, if the driver strikes a tree, it would not be his average speed that determines his survival but his speed at the instant of the collision. Similarly, when a bullet strikes a target, it is not the average speed that is significant but its instantaneous speed at the moment it strikes. So here we have distinct kinds of speeds, average speed and instantaneous speed.

The average speed of an object is defined as the object's displacement ∆x divided by the time interval ∆t during which the displacement occurs:

\begin{align*}\text{Average speed }= v=\frac {\triangle x}{\triangle t}=\frac {x_1-x_0}{t_1-t_0}\end{align*}

Notice that the points(t0, x0) and(t1, x1) lie on the position versus time curve, as the figure below shows.

This expression is also the expression for the slope of a secant line connecting the two points. Thus we conclude that the average velocity of an object between time t0 and t1 is respresented geometrically by the slope of the secant line connecting the two points (to, xo) and (t1, x1). If we choose t1 close to to, then the average velocity will closely approximate the instantaneous velocity at time to.

Geometrically, the average rate of change is represented by the slope of a secant line (figure a, below) and the instantaneous rate of change is represented by the slope of the tangent line (figure b, below).

Average Rate of Change(such as the average velocity) The average rate of change of y=f(x) over the time interval [x0, x1] is the slope msec of the secant line to the points (xo, f(x0)) and (x1, f(x0)) on the graph (figure a):

\begin{align*}m_{sec}= \frac {f(x_1)-f(x_0)}{x_1-x_0}\end{align*}

Instantaneous Rate of Change The instantaneous rate of change of y=f(x) at the point x0 is the slope msec of the tangent line to the point x0on the graph (figure b):

\begin{align*}m_{tan} = f'(x_0)= \lim_{x_1 \to x_0} \frac {f(x_1)-f(x_0)}{t_1-t_0}\end{align*}

Example 4: Suppose that y = x2 − 3.

1. Find the average rate of change of y with respect x over the interval [0, 2].
2. Find the instantaneous rate of change of y with respect x at the point x = −1.

Solution:

1. Applying the formula above for secant with f(x)=x2-3 and x0=0 and x1=2, yields
\begin{align*}m_{sec}\end{align*} \begin{align*}= \frac {f(x_1)-f(x_0)}{t_1-t_0}\end{align*}
\begin{align*}= \frac {f(2)-f(0)}{2-0}\end{align*}
\begin{align*}= \frac {1-(-3)}{2}\end{align*}
\begin{align*}= 2\end{align*}
This means that the average rate of change of y is 2 units per unit increase in x over the interval [0,2].
2. From Example 2 above, we found that f'(x) = 2x, so
mtan = f'(x0)
= f'(-1)
= 2(-1)
= -2
This means that the instantaneous rate of change is negative. That is, y is decreasing at the point x = -1. It is decreasing at a rate of 2 units per unit increase in x.

Lesson Summary

In this lesson we found a formula for the slope of a tangent line that relies on the concept of a limit. From this slope we were able to write the equation of a tangent using the point-slope form of a line. We will consider the slope of the tangent line further in the next lesson. We also considered the difference between average rate of change and instantaneous rate of change. We found that average rate of change uses secant lines and instantaneous rate of change uses tangent lines.

Points to Consider

1. What are some examples of times you would be interested in knowing the instantaneous rate of change as opposed to the average rate of change?
2. What are some examples of times you would be interested in knowing the average rate of change as opposed to the instantaneous rate of change?

Review Questions

1. Given the function y = (1/2) x2 and the values of x0 = 3 and x1 = 4, find (a) The average rate of change of y with respect to x over the interval [x0, x1]. (b) The instantaneous rate of change of y with respect to x at x0. (c) The slope of the tangent line at x1. (d) The slope of the secant line connecting x0 and x1.
2. Repeat Problem (1) for f(x) = (1\x) and the values x0 = 2 and x1 = 3.
3. Find the slope of the graph f(x) = x2 + 1 at a general point x. What is the slope of the tangent line at x0 = 6?
4. Suppose that \begin{align*}y = 1/\sqrt{x}\end{align*}. (a) Find the average rate of change of y with respect to x over the interval [1, 3]. (b) Find the instantaneous rate of change of y with respect to x at point x = 1.
5. A rocket is propelled upwards and reaches a height of h(t) = 4.9t2 in t seconds. (a) How high does it reach in 35 seconds? (b) What is the average velocity of the rocket during the first 35 seconds? (c) What is the average velocity of the rocket during the first 200 meters? (d) What is the instantaneous velocity of the rocket at the end of the 35 seconds?
6. A particle moves in the positive direction along a straight line so that after t nanoseconds, its traversed distance is given by \begin{align*}\chi (t) = 9.9 t^3\end{align*} nanometers. (a) What is the average velocity of the particle during the first 2 nanoseconds? (b) What is the instantaneous velocity of the particle at t = 2 nanoseconds.

1. (a) \begin{align*}\frac{7} {2}\end{align*}, (b) \begin{align*}3\end{align*}, (c) \begin{align*}4\end{align*}, (d) \begin{align*}\frac{7} {2}\end{align*}
2. (a) \begin{align*}\frac{-1} {6}\end{align*}, (b) \begin{align*}\frac{-1} {4},\end{align*} (c) \begin{align*}\frac{-1} {9},\end{align*} (d) \begin{align*}\frac{-1} {6}\end{align*}
3. 2x + 1, 13.
4. (a) \begin{align*}\left (\frac{\sqrt{3}} {6} - \frac{1} {2}\right )\end{align*}, (b) \begin{align*}\frac{-1} {2}\end{align*}
5. (a) 6002.5 m, (b) 171.5 m/sec, (c) 31.3 m/sec, (d) 343 m/sec
6. (a) 39.6 m/sec, (b) 118.8 m/sec

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