Net Change Theorem

 

NET CHANGE THEOREM

 

Definition:

The integral of a rate of change is the net change

 

Notes:

I. If V(t) is the volume of water in a reservoir at time t, then its derivative V'(t) is the rate at which water flows into the reservoir at time t. So

is the change in the amount of water in the reservoir between time t₁ and time t₂.

2. If the mass of a rod measured from the left end to a point x is m(x), then the linear density is ρ(x) = m' (x). So

3. If [C](t) is the concentration of the product of a chemical reaction at time t, then the rate of reaction is the derivative d[C]/dt. So

the change in the concentration of C from time t₁ to time t₂.

4. If C(x) is the cost of producing x units of a commodity, then the marginal cost is the derivative C'(x). So

is the increase is cost when production is increased from x₁ units to x₂ units

5. If the rate of growth of a population is dn/dt, then

is the net change in population during the time period from t₁ to t₂. (The population increases when births happen and decreases when death occur. The net change takes into account both births and deaths)

6. If an object moves along a straight line with position function s(t), then its velocity is v(t) = s'(t) so

is the net change of position, or displacement, of the particle during the line period from t₁ to t₂.

7. If we want to calculate the distance the object travels during the time interval, we have to consider the intervals when v(t) ≥ 0 (the particle moves to the right) and also the intervals when v(t) ≤ 0 (the particle 'moves to the left). In both cases the distance is computed by integrating |v(t)|, the speed. Therefore

 _t2∫^t2|v(t)|dt = total distance travelled

8. The acceleration of the object is a(t) = v'(t) so

is the change in velocity from time t₁ to time t₂.

 

 

PROBLEMS UNDER NET CHANGE THEOREM

 

Example 12. A particle moves along a line so that its velocity at time t is v(t) = 3t − 5, 0 ≤ t≤ 3 (measured in meters per second). Find (1) Net Change of position (2) total distance travelled.

Solution: (1) We know that,

If an object moves along a straight line with position function s(t), then its velocity is v(t) = s'(t) so

is the net change of position, or displacement, of the particle during the time period from t₁ to t₂

Here, the net change of position is

= ‒1.5 m

This means that the particle moved 1.5 m toward left.

(2) To find the total distance travelled

Given: v(t) = 3t‒5 on 0 ≤ t≤3.

Let v(t) = 0

3t‒5= 0;

t = 5/3

∴the velocity is negative on [0,5/3) and positive on (5/3,3]

The total distance travelled is given by

= 41/6 = 6.83

 

Example 13. A particle moves along a line so that its velocity at time t is v(t) = t² ‒ 2t ‒8, 1 ≤t ≤6 (measured in meters per second). Find (1) Net Change of position (2) total distance travelled.

Solution: (1) We know that,

If an object moves along a straight line with position function s(t), then its velocity is v(t) = s' (t) so

is the net change of position, or displacement, of the particle during the time period from t₁ to t₂

Here, the net change of position is

This means that the particle moved 3.33 m toward left.

(2) Given: v(t) = t²‒2t‒8 = (t ‒ 4)(t + 2) on 1 ≤ t≤ 6.

Let v(t) = 0

 (t‒4)(t+2) = 0;

 t = 4 or ‒2

 t = ‒2 is not possible.

∴ the velocity is negative on [1,4) and positive on (4, 6].

The total distance travelled is given by

= 32.67

 

Example 14. A particle moves along a line so that its velocity at time t is v(t) = t² − t − 6 (measured in meters per second)

(1) Find the displacement of the particle during the time period 1 ≤ t ≤ 4

(2) Find the distance traveled during this time period

Solution: We know that,

If an object moves along a straight line with position function s(t), then its velocity is v(t) = s'(t) so

is the net change of position, or displacement, of the particle during the time period from t₁ to t₂

(1) Here, the displacement is

 = ‒45

This means that the particle moved 4.5 m toward left.

(2) Given: v(t) = t²‒t‒6 = (t‒3)(t+2) on 1 ≤ t ≤ 4.

Let v(t) = 0

 (t − 3)(t + 2) = 0;

  t =3 or 2

 t = ‒2 is not possible.

the velocity is negative on [1,3) and positive on (3,4].

The total distance travelled is given by

= 10.17 m.

 

Example 15. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed. Suppose Andrew takes his iceboat out one morning when a light ‒mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the wind speed increases according to the function v(t) = 20t + 5. For the second half hour of Andrew's outing, the wind remains steady at 15 mph. In other words, the wind speed is given by

Recalling that Andrew's iceboat travels at twice the wind speed, and assuming he moves in a straight line away from his starting point, how far is Andrew from his starting point after 1 hour?

Solution: To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then

Distance = ₀∫¹2v(t)dt.

Substituting the expressions we were given for v(t), we get

= [20/4 + 5] ‒ 0 + [30 ‒ 15]

= 25

Andrew is 25 mi from his starting point after 1 hour.

 

Example 16. The acceleration function (in m/s²) and the initial velocity are given for a particle moving along a line. Find

(1) the velocity at time t and

(2) the distance travelled during the given time interval a(t) = t + 4, v(0) = 5,0 ≤ t ≤ 10

Solution: (1) If a(t) = t + 4 is acceleration and v(0) given, then

v(t) = v(0) + ₀ʃ⁴a(t) dt     ... (1)

and position change (displacement) on [0,T) is

s = ₀∫^T v(t) dt (So split where v(0) = 0)

Check sign of v in 0 ≤ t ≤ 10

Solve: t²/2 + 4t + 5 = 0

⇒ t = −4±√6 both negative.

So, v(t) > 0 in 0 ≤ t≤ 10

(b) The distance travelled during the time interval [0,10] is given by

= 1250/3 m

So, the distance traveled = 1250/3 m ≈ 416.67m

 

Example 17. The acceleration function (in m/s²) and the initial velocity are given for a particle moving along a line. Find

(a) the velocity at time t and

(b) the distance traveled during the given time interval a(t) = 2t+3, v(0) = − 4, 0 ≤ t ≤ 3

Solution: (a) If a(t) = 2t + 3 is acceleration and v(0) given, then

v(t) = v(0) + ₀∫^t a(s) ds

(b) Check sign of v in 0 ≤ t ≤ 3

Solve: t²+3t‒4 = 0

t = 1,‒4

So, v(t) < 0 on [0,1)

v(t)> 0 on (1,3]

S = 0∫3 v(t) dt

So, the distance traveled= 89/6 m ≈ 14.83 m

 

Example 18. The linear density of a rod of length 4m is given by p(x) = 9 + 2√x measured in kilograms per meter, where x is measured in meters from one end of the rod. Find the total mass of the rod.

Solution:

The Net change theorem says

Total change in quantity = _a∫^b rate of change dx

Here the quantity is mass, the rate of change is linear density ρ(x) in kg/m.

So mass = ₀∫⁴ ρ(x) dx

 

Example 19. Water flows from the bottom of a storage tank at a rate of r(t) = 200‒4t liters per minute, 0 ≤ t ≤ 50. Find the amount of water that flows from the tank during the first 10 minutes.

Solution:

Given: Rate of flow r(t) = 200 ‒ 4t, 0 ≤ t ≤ 10

Net change theorem

Amount of water out flow = ₀∫¹⁰ r(t) dt

 1800 litres of water flows out in the first 10 minutes.

 

Example 20. The marginal cost of manufacturing x meters of a certain fabric is C'(x) = 3 ‒0.01x + 0.000006x² (in dollars per meter). Find the increase in cost is the production level is raised from 2000 meters to 4000 meters.

Solution:

Given: C'(x) = 3 ‒ 0.01x + 0.000006x²

We want to increase in cost

Find an antiderivative

F(x) = 3 ‒ 0.05x² + 0.000002x³

F(4000) = 3(4000) ‒ 0.005(4000)² + 0.000002(4000)³

= 60,000

F(2000) = 3(2000) ‒ 0.005(2000)² + 0.000002(2000)³

= 2000

 ∆C = F(4000) – F(2000)

= 60000 ‒ 2000 = 58000

The cost increases by $58,000

 

 

EXERCISE

 

8. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed. Suppose Andrew takes his iceboat out one morning when a light ‒mph breeze has been blowing all morning. As Andrew gets his iceboat set up, though, the wind begins to pick up. The wind speed is given by

Under these conditions, how far from his starting point is Andrew after 1 hour?

Ans: 17.5 mi

 

9. Find the net displacement and total distance traveled in meters given the velocity function f(t) = ½ e^t ‒ 2 over the interval [0,2].

Ans: ‒0.8055 m, 1.74 m

 

 

 

PROBLEMS UNDER INDEFINITE INTEGRALS INTEGRATION OF SOME STANDARD FUNCTIONS

 

(i) ∫dx = x + c

(ii) ∫ xⁿ dx = (xⁿ⁺¹ / n+1) + c   (n≠‒1)

(iii) ∫ dx/x = logx + c

(iv) ∫ e^x dx = e^x + c

(v) ∫ a^x dx = a^x loga + c

(vi) ∫ sinx dx = −cosx + c

(vii) ∫ cosx dx = sinx + c

(viii) ∫ sec²x dx = tan x + c

(ix) ∫ secx tanx dx = secx + c

(x) ∫ cosec² x dx = ‒ cot x + c

(xi) ∫ cosecx cotx dx = ‒cosec x + c

(xii) ∫ tanx dx = log secx + c

(xiii) ∫ secx dx = log |secx + tan x| + c

(xiv) ∫ cotx dx = log sin x + c

(xv) ∫ cosec x dx = log |cosecx ‒ cotx| + c

(xvi) ∫ dx / [1+x²] = tan^‒1x + c

(xvii) ∫ dx / √[1‒x²] = sin^‒1x + c

 

Example 21. Evaluate ʃ (10x⁴ ‒ 3sec²x)dx

Solution: Let I = ∫(10x⁴ ‒ 3sec²x) dx

= 10∫x⁴dx ‒3∫sec²xdx

= 10 (x⁵/5) ‒ 3tanx + c

 I = 2x⁵ ‒ 3tanx + c

 

Example 22. Evaluate ∫ ( [x³ ‒ 2√x] / x ) dx

Solution:

 

Example 23. Evaluates, ₁∫⁹ ( [2t² + t²√t ‒ 1] / t² ) dt

 

Example 24. Evaluate ∫ ( cosθ / sin²θ ) dθ

 = secx + log(cosec x ‒ cot x)

 

 

EXERCISE

 

10. Evaluate the following