Lesson 14 Tangent Lines and Rates of Change

September 20, 2013

Lesson 14 Tangent Lines and Rates of Change

1 / 12

Announcements

Please check your e-mail for the Informal Early Feedback survey!

Lesson 14 Tangent Lines and Rates of Change

2 / 12

Homework 13 Questions?

Lesson 14 Tangent Lines and Rates of Change

3 / 12

Overview Since the last exam, we’ve covered: 1. Section 2.2 – Power rule.

Today we’ll finish up section 2.2 by reviewing what we learned about tangent lines before the first exam. We’ll also study a few word problems about rates of change. Recall that the equation of the line tangent to the graph of f (x) when x = a is given by the formula, y = f 0 (a)(x − a) + f (a). The Connect HW will always want this in the simplified form y = mx + b.

Lesson 14 Tangent Lines and Rates of Change

4 / 12

Example 1 (HW #1–5.) Find the equation of the line tangent to the graph of x 3 +

Lesson 14 Tangent Lines and Rates of Change

5 / 12

√

x at x = 4.

Example 1 (HW #1–5.) Find the equation of the line tangent to the graph of x 3 + The equation of the line is, according to the formula, y = f 0 (4)(x − 4) + f (4).

Lesson 14 Tangent Lines and Rates of Change

5 / 12

√

x at x = 4.

Example 1 (HW #1–5.) Find the equation of the line tangent to the graph of x 3 + The equation of the line is, according to the formula, y = f 0 (4)(x − 4) + f (4). We have, f (4) = 43 +

√

4 = 66

Lesson 14 Tangent Lines and Rates of Change

5 / 12

√

x at x = 4.

Example 1 (HW #1–5.) Find the equation of the line tangent to the graph of x 3 +

√

The equation of the line is, according to the formula, y = f 0 (4)(x − 4) + f (4). We have, f (4) = 43 +

√

4 = 66

and d 3 √ 1 2 f (4) = [x + x] = 3x + 1/2 dx 2x x=4 x=4 0

1 1 193 = 3 · 42 + √ = 48 + = 4 4 2 4

Lesson 14 Tangent Lines and Rates of Change

5 / 12

x at x = 4.

Example 1, con’t Since f (4) = 66 and f 0 (4) =

183 , we have 4

y=

193 (x − 4) + 66. 4

Simplified, this is y=

193 x − 127. 4

Lesson 14 Tangent Lines and Rates of Change

6 / 12

Example 1, con’t Since f (4) = 66 and f 0 (4) =

183 , we have 4

y=

193 (x − 4) + 66. 4

Simplified, this is y=

193 x − 127. 4

Warning: It’s important to take the derivative before evaluating it. Otherwise, by the constant rule d d f (4) = 66 = 0, dx dx which is not helpful. Lesson 14 Tangent Lines and Rates of Change

6 / 12

Rates of change Recall that earlier we said the derivative can be interpreted as either the slope of the tangent line or as the instantaneous rate of change. In the word problems of this section, you’ll need to be able to distinguish between the rate of change f 0 and the actual function f .

Lesson 14 Tangent Lines and Rates of Change

7 / 12

Rates of change Recall that earlier we said the derivative can be interpreted as either the slope of the tangent line or as the instantaneous rate of change. In the word problems of this section, you’ll need to be able to distinguish between the rate of change f 0 and the actual function f . For example, let’s say that a population model for a group of cardinals predicts that after t months, the population will be P(t) = −0.5t 2 + 56t + 43 thousand birds.

Lesson 14 Tangent Lines and Rates of Change

7 / 12

Rates of change Recall that earlier we said the derivative can be interpreted as either the slope of the tangent line or as the instantaneous rate of change. In the word problems of this section, you’ll need to be able to distinguish between the rate of change f 0 and the actual function f . For example, let’s say that a population model for a group of cardinals predicts that after t months, the population will be P(t) = −0.5t 2 + 56t + 43 thousand birds. Then the growth rate of the population after three months is P 0 (3).

Lesson 14 Tangent Lines and Rates of Change

7 / 12

Rates of change Recall that earlier we said the derivative can be interpreted as either the slope of the tangent line or as the instantaneous rate of change. In the word problems of this section, you’ll need to be able to distinguish between the rate of change f 0 and the actual function f . For example, let’s say that a population model for a group of cardinals predicts that after t months, the population will be P(t) = −0.5t 2 + 56t + 43 thousand birds. Then the growth rate of the population after three months is P 0 (3). The net number of birds born in the third month, however, is P(3) − P(2). Lesson 14 Tangent Lines and Rates of Change

7 / 12

Example 2 (HW #6–12.) Using an asset allocation model he got off the radio, George expects to earn P(t) = 0.5t 3 − 1.5t 2 + t thousand dollars after t months. How fast is the investment growing after 2 months?

Lesson 14 Tangent Lines and Rates of Change

8 / 12

Example 2 (HW #6–12.) Using an asset allocation model he got off the radio, George expects to earn P(t) = 0.5t 3 − 1.5t 2 + t thousand dollars after t months. How fast is the investment growing after 2 months? What we need to calculate is P 0 (2). By the power rule, P 0 (t) = 1.5t 2 − 3t + 1 So P 0 (2) = 1.5 · 22 − 3 · 2 + 1 = 1, however, the expected growth rate is $1,000/month.

Lesson 14 Tangent Lines and Rates of Change

8 / 12

Example 2, part 2 Using an asset allocation model he got off the radio, George expects to earn P(t) = 0.5t 3 − 1.5t 2 + t thousand dollars after t months. What was the investment’s net growth between the first and third month?

Lesson 14 Tangent Lines and Rates of Change

9 / 12

Example 2, part 2 Using an asset allocation model he got off the radio, George expects to earn P(t) = 0.5t 3 − 1.5t 2 + t thousand dollars after t months. What was the investment’s net growth between the first and third month? What we need to calculate is P(3) − P(1). No derivatives needed: P(3) − P(1) = [0.5 · 33 − 1.5 · 32 + 3] − [0.5 · 13 − 1.5 · 12 + 1] = 3 − 0 = $3,000.

Lesson 14 Tangent Lines and Rates of Change

9 / 12

Example 3 If Audi spends x million dollars on advertising their cars in a month, they 50 35 − 2 million expect their profit for that month to be P(x) = 80 + x x dollars. How are sales changing when $6 million is spent on advertising?

Lesson 14 Tangent Lines and Rates of Change

10 / 12

Example 3 If Audi spends x million dollars on advertising their cars in a month, they 50 35 − 2 million expect their profit for that month to be P(x) = 80 + x x dollars. How are sales changing when $6 million is spent on advertising? We seek P 0 (6), which is   d 50 35 50 70 − 2 80 + = − 2 + 3 ' −1.065. dx x x x x x=6 x=6 Therefore profits are decreasing at a rate of 1.065 million per additional million spent on advertising. Audi should probably cut back on their marketing department. Lesson 14 Tangent Lines and Rates of Change

10 / 12

Review Today: 1. The slope of a tangent line, an instantaneous rate of change, and so on, are all found by calculating the derivative. 2. The actual change in a function between two points is found by calculating the difference of the two values of the function. 3. The average change is a difference quotient. (Not appearing on this homework). For Monday: 1. Read p. 132–137 on the Product Rule and the Quotient Rule. 2. Example 2.3.6 is critical – avoid using the quotient rule whenever possible. Get ready for the quiz! Lesson 14 Tangent Lines and Rates of Change

11 / 12

Quiz 5 (15 minutes) Find the derivative. 1. −2x 5 . 2. 3x 2/3 .

3. −

3 . x2

2 4. − √ . x

5.

√ 5

x 8. Lesson 14 Tangent Lines and Rates of Change

12 / 12