MATH 12002 - CALCULUS I 4.3: The Net Change Theorem Professor - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §4.3: The Net Change Theorem

Professor Donald L. White

Department of Mathematical Sciences Kent State University

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Net Change Theorem

If F(x) is an antiderivative for f (x) on the interval [a, b], then f represents the rate of change of F and by the Evaluation Theorem, b

a f (x) dx = F(b) − F(a), which is the net change in F on the interval.

Hence we have:

Net Change Theorem

If F is differentiable on [a, b] with rate of change f , then the net change in F on the interval [a, b] is F(b) − F(a) = b

a

f (x) dx.

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Displacement & Total Distance

A familiar special case of the Net Change Theorem involves position s(t) and its rate of change v(t). We have seen that b

a

v(t) dt = s(b) − s(a), the displacement on the interval from time t = a to time t = b. This is the (signed) distance between the starting position and the ending position; it does not account for all of the distance that may have been travelled during the time period. In fact, displacement is the distance travelled in the positive direction (i.e., positive velocity) minus the distance travelled in the negative direction (i.e., negative velocity).

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Displacement & Total Distance

To find the total distance travelled, we need the distance travelled in the positive direction (i.e., positive velocity) plus the distance travelled in the negative direction (i.e., negative velocity). Similar to our computation of area in terms of integrals, Total Distance Travelled = the sum of integrals of v(t)

• n intervals where v(t) is positive

minus the sum of integrals of v(t)

• n intervals where v(t) is negative

= b

a

|v(t)| dt. (See text page 229 for a more detailed explanation.)

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Example

Example

Find the displacement and total distance travelled on the interval t = 0 to t = 5 when the velocity function is v(t) = t2 − 6t + 8. Displacement: 5 v(t) dt = 5 (t2 − 6t + 8) dt = 1

3t3 − 3t2 + 8t

• 5

= 1

3(53) − 3(52) + 8(5)

• −

1

3(03) − 3(02) + 8(0)

• =

125 3 − 75 + 40 = 20 3 . Hence displacement on the interval is 20 3 .

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Example

Total Distance: We know that total distance is 5

0 |v(t)| dt and

|v(t)| =

• v(t)

if v(t) 0, −v(t) if v(t) 0. We need to determine when v(t) = t2 − 6t + 8 is positive and when it is negative. Notice that v(t) = t2 − 6t + 8 = (t − 2)(t − 4), so v(t) can only change sign at t = 2 and at t = 4. We have v(0) = (−2)(−4) is positive, v(3) = (1)(−1) is negative, and v(5) = (3)(1) is positive. Therefore, v(t) 0 on [0, 2] and [4, 5], and v(t) 0 on [2, 4], and so

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Example

5 |v(t)| dt = 5 |t2 − 6t + 8| dt = 2 |t2 − 6t + 8| dt + 4

2

|t2 − 6t + 8| dt + 5

4

|t2 − 6t + 8| dt = 2 t2 − 6t + 8 dt + 4

2

−(t2 − 6t + 8) dt + 5

4

t2 − 6t + 8 dt = 2 t2 − 6t + 8 dt + 4

2

−t2 + 6t − 8 dt + 5

4

t2 − 6t + 8 dt = 20 3 + 4 3 + 4 3 = 28 3 .

Hence total distance travelled on the interval is 28 3 .

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