Tag Archives: O Levels

Further applications of differentiation 1

The diagram shows a river,500m wide between the straight parallel banks where AD = 500m and   DB = 2000. A man swims across the river from A to C at 0.5 m/s and then he runs along the banks from C to B at 0.8 m/s

(a)If he swims in the direction AC making an angle of Ɵ and reaches C in t seconds, show that t = 1000cosec Ɵ

(b)If T(in seconds) is the total time taken from A to B, express T in terms of Ɵ

(c)Find the minimum value of T in minutes and seconds, correct to the nearest second.

 

(a)

Using trigo ratio,

sin Ɵ = 500/AC

AC = 500/sin Ɵ

 = 500cosec Ɵ

 0.5 = 500cosec Ɵ/t

 T = 1000cosec Ɵ

 

(b)

BC = 2000 – DC

BC = 2000 – 500/tan Ɵ

TBC = BC/s = 2500 – 625/tan Ɵ

Total time, T = tAC + tBC

= 1000cosec Ɵ + 2500 – 625/tan Ɵ

 

(c)

 

Equations without stationary points

Given y = 4/√(2x-3), x > 1.5

(a)Find dy/dx

(b)Explain why the graph of y = 4/√(2x – 3) does not have a stationary point

(c)State whether y = 4√(2x – 3) is an increasing or decreasing function. Explain your answer clearly.

y = 4(2x – 3)-1/2

dy/dx = -2(2x – 3)-3/2(2)

           = -4(2x – 3)-3/2

(b) Since, dy/dx = -4(2x – 3)-3/2  , (2x-3) is always positive

     (2x – 3 )-3/2 > 0

  • dy/dx < 0, dy/dx ≠ 0, therefore graph has no stationary point

Maximum/Minimum gradient of a curve

Find the minimum gradient of the curve y = 2x3 – 9x2 + 5x + 3 and the value of x when the minimum gradient occurs.

y = 2x3 – 9x2 + 5x + 3

gradient, m = 6x2 – 18x + 5
dm/dx = 12x – 18

minimum gradient,

dm/dx = 0

12x – 18 =0

X = 3/2 occurs at minimum gradient

= 6(3/2)2 – 18(3/2) + 5

= -17/2

Maximum point and minimum point 1

A curve with equation in the form of y = ax + b/x 

y = ax + b/x2 has a stationary point at (3 , 4), where a and b are constants. Find the value of a and of b.

X = 3     y = 4

4 = 3a + b/a —–(1)

Y = ax + bx-2

dy/dx = a – 2bx-3—–(2)

sub dy/dx = 0, x = 3 into (2)

0 = a – 2b(3)-3

0 = a – 2b/27—–(3)

Sub (3) into (1)

4 = 3(2b/27) + b/9

4 = 1/3b

b = 12  ,   a = 2(12)/27

b = 12  ,   a = 8/9

Solving integration by the use of partial fractions

Solution:

(a) A=2, B=-5, C=0

(b) 2x/(x2+3)

(c) ln(2x-1) – (5/2)ln(x2+3) + C

(a)6 + 5x – 8x2 = A(x2 + 3) + (Bx + c)(2x – 1)

Sub x = ½

6 + 5/2 – 8(1/2)2 = A(1/4 + 3) + 0

13/2 = (13/4)A —— A=2

When x=0, 6 = 2(3) + c(-1) —— C=0

Compare coefficients of x2

-8 = A + 2B

-10 – 2B —— B = -5

 

(b)d/dx[ln(x2+3)] = 2x/(x2+3)

=2x/(x2+3)

 

(c)∫ 2/(2x-1) + ∫(-5x)/(x2+3)

=ln(2x-1) – (5/2)ln(x2+3) + C