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Learn/ 11th Class Mathematics/ Unit 13 /Numericals

Unit 13: Differentiation — Exercise 13 3

11th Class Mathematics · Unit 13: Differentiation

1.A car's position at time t is given by s(t) = 5t^3 - 3t^2 + t. Find the velocity by differentiating the position function with respect to time.
Given
Position function s(t) = 5t3 - 3t2 + t
Formula
Differentiate w.r.t t ddts(t) = 5ddtt3 - 3ddtt2 + ddt(t)
s'(t) = 15t2 - 6t + 1
Result
Velocity V = 15t2 - 6t + 1
2.Structural stress on a bridge is modeled by function S(x) = 100 - 5x^2, where x is the distance from the center of the bridge. (a) Find the point where stress is maximum. (b) Calculate the rate of change at that point.
Given
Stress function S(x) = 100 - 5x2
Formula
(a) Differentiate w.r.t x S'(x) = -10x
(a) For maximum stress, set S'(x)=0 0 = -10x ⇒ x = 0
Working
(a) Substituting x=0 into S(x) S(0) = 100 - 5(0)2 = 100
Result
(a) Maximum-stress point (0, 100)
(b) Rate of change at x=0 S'(0) = 10(0) = 0
Sign dropped: should read S'(0) = -10(0) = 0. The numeric result (0) is unaffected since the term is multiplied by 0.
3.A company's revenue function is given by R(x) = 1000x - 10x^2, where x is the number of units produced. The cost function is C(x) = 300x + 2000. (i) Find the profit function P(x). (ii) Determine the marginal profit when x = 15.
Formula
(i) Profit = Revenue - Cost P(x) = R(x) - C(x)
Working
(i) Substituting R(x) and C(x) P(x) = 1000x - 10x2 - 300x - 2000
Result
(i) Profit function P(x) = -10x2 + 700x - 2000
Working
(ii) Profit at x=15 P(15) = -10(15)2 + 700(15) - 2000
Marginal profit means dP/dx, not P(x) itself. The correct approach is P'(x) = -20x + 700, so P'(15) = -300 + 700 = 400.
P(15) = -2250 + 10500 - 2000
This sum equals 6250, not 400. The printed final answer of Rs. 400 actually matches the marginal profit P'(15), not this (incorrect) evaluation of P(15).
Result
(ii) Marginal profit P(15) = Rs. 400
4.An investment grows according to the function A(t) = 10000(1+0.05t)^3, where A(t) is the value of the investment and t is the time in years. (i) Find the rate of change of the investment after 8 years. (ii) What is the investment value after 8 years?
Given
Investment function A(t) = 10000(1+0.05t)3
Formula
(i) Differentiate w.r.t t ddtA(t) = ddtleft[10000(1+0.05t)3right]
A'(t) = 30000(1+0.05t)2(0.05) = 1500(1+0.05t)2
Working
(i) At t=8 A'(8) = 1500(1+0.05 × 8)2 = 1500(1.4)2
Result
(i) Rate of change A'(8) = 1500 × 1.96 = 2940
Working
(ii) At t=8 A(8) = 10000(1+0.05 × 8)3 = 10000(1.4)3
Result
(ii) Investment value A(8) = 27440
5.The position of a particle moving along a line is given by s(t) = 5t^3 - 12t^2 + 8t, where s(t) is the position in meters and t is the time in seconds. (i) Determine the velocity of the particle at t=4 seconds. (ii) Find the acceleration at t=4 seconds. (iii) When is the particle at rest?
Given
Position function s(t) = 5t3 - 12t2 + 8t
Formula
(i) Differentiate w.r.t time s'(t) = 15t2 - 24t + 8
Working
(i) At t=4 sec V|t=4 = 15(4)2 - 24(4) + 8
Result
(i) Velocity V = 152 m/s
Formula
(ii) Differentiate velocity w.r.t time dvdt = 15ddtt2 - 24ddt(t) + ddt(8)
a = 30t - 24
Working
(ii) At t=4 sec a = 30 × 4 - 24 = 120 - 24
Result
(ii) Acceleration a = 96 m/s2
Formula
(iii) Set velocity to zero 15t2 - 24t + 8 = 0
(iii) By quadratic formula t = frac{24 ± sqrt{(24)2-4(15)(8)}{2(15)} = frac{24 ± sqrt{96}{30}
Result
(iii) Times at rest t = 1.13 s or t = 0.47 s
6.The position of a car traveling along a straight highway is given by x(t) = 30t^2 - 4t^3, where x(t) is the distance traveled in kilometers and t is the time in hours. (i) Find the car's velocity at t=3 hours. (ii) Determine the car's acceleration at t=3 hours.
Given
Position function x(t) = 30t2 - 4t3
Formula
(i) Differentiate w.r.t time ddt[x(t)] = 30ddt(t2) - 4ddt(t3)
x'(t) = 60t - 12t2
Working
(i) At t=3 V(3) = 60(3) - 12(3)2
Result
(i) Velocity V(3) = 72 km/h
Formula
(ii) Differentiate velocity w.r.t t ddt[V(t)] = 60ddt(t) - 12ddt(t2)
V'(t) = 60 - 24t
Working
(ii) At t=3 a(3) = 60 - 24(3)
Result
(ii) Acceleration a(3) = -12 km/h2 (deceleration)
7.The stress on a beam under a varying load is given by S(x) = 400x - x^3, where S(x) is the stress in pascals Pa and x is the distance from the fixed end in meters. Calculate the rate of change of stress at 6 meters.
Given
Stress function S(x) = 400x - x3
Formula
Differentiate w.r.t x ddx[S(x)] = 400ddx(x) - ddx(x3)
S'(x) = 400 - 3x2
Working
At x=6 m S'(6) = 400 - 3(6)2
Result
Rate of change of stress S'(6) = 292 Pa/m
8.The cost C(r) to construct a cylindrical tank depends on the radius of the base, and is given by C(r) = 8000\pi r^2 + 150000/r, where the first term represents the cost of the base and the second term represents the cost of the walls. Determine the rate of change of the cost at r=4 meters.
Given
Cost function C(r) = 8000pi r2 + 150000r
Formula
Differentiate w.r.t r C'(r) = 16000pi r - 150000r2
Working
At r=4 m C'(4) = 16000pi (4) - 15000042
Result
Rate of change of cost C'(4) = 191686.93 units/m