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2.1: Tangent Lines and Rates of Change

Created by: CK-12

Learning Objectives

A student will be able to:

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

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 is it related to the tangent lines on a curve.

The Tangent Line

If two points P(x_0, y_0) and Q(x_1, y_1) are two different points of the curve y = f(x) (Figure 1), then the slope of the secant line connecting the two points is given by

 m_{sec} = \frac{y_1 - y_0} {x_1 - x_0} = \frac{f(x_1) - f(x_0)} {x_1 - x_0}& &(1)

Now if we let the the point x_1 approach x_0, Q will approach P along the graph f. Thus the slope of the secant line will gradually approach the slope of the tangent line as x_1 approaches x_0. Therefore (1) becomes

 m_{tan} =\lim_{x_1 \to x_0} \frac{f(x_1) - f(x_0)} {x_1 - x_0}.& &(2)

If we let h = x_1 - x_0, then x_1 = x_0 + h and h \rightarrow 0 becomes equivalent to x_1 \rightarrow x_0, so (2) becomes

 m_{tan} =\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)} {h}.

If the point P(x_0, y_0) is on the curve f, then the tangent line at P has a slope that is given by

 m_{tan} =\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)} {h}

provided that the limit exists.

Recall from algebra that the point-slope form for the tangent line is given by

y - y_0 = m_{tan}(x - x_0).

Example 1:

Find the slope of the tangent line to the curve f(x) = x^3 passing through point P(2, 8).

Solution:

Since P(x_0, y_0) = (2, 8), using the slope of the tangent equation,

 m_{tan} =\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)} {h}

we get

m_{tan} &=\lim_{h \to 0} \frac{f(2 + h) - f(2)} {h}\\&=\lim_{h \to 0} \frac{(h^3 + 6h^2 + 12h + 8) - 8} {h}\\&=\lim_{h \to 0} \frac{h^3 + 6h^2 + 12h} {h}\\&=\lim_{h \to 0} (h^2 + 6h + 12)\\&= 12.

Thus the slope of the tangent line is 12. Using the point-slope formula above,

y - 8 = 12(x - 2)

or

y = 12x - 16

Next we are interested in finding a formula for 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 replace the constant x_0 by the variable x. This yields

 m_{tan} =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h}.

We denote this formula by

 f'(x) =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h},

where f(x) is read "\mathit{f} prime of \mathit{x}." The next example illustrate its usefulness.

Example 2:

If f(x) = x^2 - 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) =\lim_{h \to 0} \frac{f(x + h) - f(x)} {h},

then

f'(x) &=\lim_{h \to 0} \frac{[(x + h)^2 - 3] - [x^2 - 3]} {h}\\&=\lim_{h \to 0} \frac{x^2 + 2xh + h^2 - 3 - x^2 + 3} {h}\\&=\lim_{h \to 0} \frac{2xh + h^2} {h}\\&=\lim_{h \to 0} (2x + h)\\&= 2x

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 slopes of the tangent lines at x = 2 and x = -1 are 4 and -2, respectively.

Example 3:

Find the slope of the tangent line to the curve y = \frac{1}{x} that passes through the point (1, 1).

Solution:

Using the slope of the tangent formula

 f'(x) =\lim_{h \to 0}\frac{f(x + h) - f(x)} {h}

and substituting y = \frac{1}{x},

y' &=\lim_{h \to 0} \frac{\left (\frac{1} {x+h} \right) - \frac{1} {x}} {h}\\&= \lim_{h \to 0} \frac{\frac{x - x - h} {x(x + h)}} {h}\\&= \lim_{h \to 0} \frac{x - x - h} {hx(x + h)}\\&= \lim_{h \to 0} \frac{-h} {hx(x + h)}\\&= \lim_{h \to 0} \frac{-1} {x(x + h)}\\&= \lim_{h \to 0} \frac{-1} {x(x + h)}\\&= \frac{-1} {x^2}.

Substituting x = 1,

y' &= \frac{-1} {1}\\&= -1.

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

y - y_0 = m(x - x_0),

where (x_{0}, y_{0}) = (1, 1).

y - 1 &= -1 ( x - 1)\\&= -x + 1 + 1\\&= -x + 2,

which is the equation of the tangent line.

Average Rates of Change

The primary concept of calculus involves calculating the rate of change of a quantity with respect to another. For example, speed is defined as the rate of change of the distance travelled with respect to time. If a person travels 120\;\mathrm{miles} in four hours, his speed is 120/4 = 30 \;\mathrm{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\;\mathrm{miles} at a rate of 30 \;\mathrm{mi/hr} for four hours does not do so continuously. He must have slowed down or sped up during the four-hour period. But it does suffice to say that he traveled for four hours at an average rate of 30\;\mathrm{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 two distinct kinds of speeds, average speed and instantaneous speed.

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

 v = \frac{\triangle x} {\triangle t} = \frac{x_1 - x_0} {t_1 - t_0}.

Notice that the points (t_0, x_0) and (t_1, x_1) lie on the position-versus-time curve, as Figure 1 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 t_0 and t_1 is represented geometrically by the slope of the secant line connecting the two points (t_0, x_0) and (t_1, x_1). If we choose t_1 close to t_0, then the average velocity will closely approximate the instantaneous velocity at time t_0.

Geometrically, the average rate of change is represented by the slope of a secant line and the instantaneous rate of change is represented by the slope of the tangent line (Figures 2 and 3).

Average Rate of Change (such as the average velocity)

The average rate of change of x = f(t) over the time interval [t_0, t_1] is the slope m_{sec} of the secant line to the points (t_0, f(t_0)) and (t_1, f(t_1)) on the graph (Figure 2).

 m_{sec} = \frac{f(t_1) - f(t_0)} {t_1 - t_0}

Instantaneous Rate of Change

The instantaneous rate of change of x = f(t) at the time t_0 is the slope m_{tan} of the tangent line at the time t_0 on the graph.

 m_{tan} = f'(t_0) = \lim_{t_1 \to t_0} \frac{f(t_1) - f(t_0)} {t_1 - t_0}

Example 4:

Suppose that y = x^2 - 3.

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

Solution:

1. Applying the formula for Average Rate of Change with f(x) = x^2 - 3 and x_0 = 0 and x_1 = 2\;\mathrm{yields}

m_{sec} &= \frac{f(x_1) - f(x_0)} {x_1 - x_0}\\&= \frac{f(2) - f(0)} {2 - 0}\\&= \frac{1 - (-3)} {2}\\&= 2

This means that the average rate of change of y is 2\;\mathrm{units \ per \ unit} increase in x over the interval [0, 2].

2. From the example above, we found that f'(x) = 2x, so

m_{tan} &= f'(x_0)\\&= f'(-1)\\&= 2(-1)\\&= -2

This means that the instantaneous rate of change is negative. That is, y is decreasing at x = -1. It is decreasing at a rate of 2\;\mathrm{units \ per \ unit} increase in x.

Multimedia Links

For a video explaining instantaneous rates of change (4.2), see Slopes of Tangents and Instantaneous Rates of Change (9:26).

For a video with an application regarding velocity (4.2), see Calculus Help: Instantaneous Rates of Change (9:03).

The following applet illustrates how the slope of a secant line can become the slope of the tangent to a curve at a point as h \to 0. Follow the directions on the page to explore how changing the distance between the two points makes the slope of the secant approach the slope of the tangent Slope at a Point Applet. Note that the function and the point of tangency can also be edited in this simulator.

Review Questions

  1. Given the function y = 1/2 \ x^2 and the values of x_0 = 3 and x_1 = 4, find
    1. The average rate of change of y with respect to x over the interval [x_0, x_1].
    2. The instantaneous rate of change of y with respect to x at x_0.
    3. The slope of the tangent line at x_1.
    4. The slope of the secant line between points x_0 and x_1.
    5. Make a sketch of y = 1/2 \ x^2 and show the secant and tangent lines at their respective points.
  2. Repeat problem #1 for f(x) = 1/x and the values x_0 = 2 and x_1 = 3.
  3. Find the slope of the graph f(x) = x^2 + 1 at a general point x. What is the slope of the tangent line at x_0 = 6?
  4. Suppose that  y = 1 / \sqrt{x}.
    1. Find the average rate of change of y with respect to x over the interval [1,3].
    2. Find the instantaneous rate of change of y with respect to x at point x = 1.
  5. A rocket is propelled upward and reaches a height (in meters) of h(t) = 4.9t^2 in t\;\mathrm{seconds}.
    1. How high does it reach in 35 \;\mathrm{seconds}?
    2. What is the average velocity of the rocket during the first 35 \;\mathrm{seconds}?
    3. What is the average velocity of the rocket during the first 200 \;\mathrm{meters}?
    4. What is the instantaneous velocity of the rocket at the end of the 35 \;\mathrm{seconds}?
  6. A particle moves in the positive direction along a straight line so that after t\;\mathrm{nanoseconds}, its traversed distance is given by  \chi(t) = 9.9t^3\;\mathrm{nanometers}.
    1. What is the average velocity of the particle during the first 2\;\mathrm{nanoseconds}?
    2. What is the instantaneous velocity of the particle at t = 2\;\mathrm{nanoseconds}?

Review Answers

    1. \frac {7}{2}
    2.  3
    3.  4
    4. \frac {7}{2}
    5. Not included
    1.  \frac {-1}{6}
    2.   \frac {-1}{4}
    3.   \frac {-1}{9}
    4.   \frac {-1}{6}
    5. Not included
  1. 2x, 12.
    1.  \left ( \frac {\sqrt{3}}{6} -  \frac{1}{2} \right )
    2.  -\frac{1}{2}
    1. 6002.5 \;\mathrm{m}
    2. 171.5 \;\mathrm{m/sec}
    3. 31.3 \;\mathrm{m/sec}
    4. 343 \;\mathrm{m/sec}
    1. 39.6 \;\mathrm{m/sec}
    2. 118.8 \;\mathrm{m/sec}

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Date Created:

Feb 23, 2012

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Aug 07, 2014
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CK.MAT.ENG.SE.1.Calculus.2.1

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