Monthly Archives: January 2018

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

CJC/II/Q4

The diagram shows the region R bounded by the two parabolas 

      Y = x2 and x = (y – 2)2 – 2  and the y – axis. Find the points indicated A and B in the diagram.

(a)Find the area of the region R

(b)Find the volume formed when R is rotated 2π radian about the y – axis

Solutions:  A(-1 , 1), B(0 , 2 – 2√2) 

                 (a) area = 0.448

                 (b) volume = 1.10

Sub Y = x2 into x = (y – 2)2 – 2

y = 2 ±√x+2

x = (x2 – 2)2 – 2

x = x4 – 4x2 + 4 – 2

0 = x4 – 4x2 – x + 2 = (x2 – x – 2)(x2 + x – 1)

X = 2, -1, -1.618, 0.6180

By long division,  Quadratic factor = (x – 2)(x + 1) = x2 – x – 2

 

A(-1 , 1)    B( (-1 + √5)/2 , [(-1 + √5)/2]2)

 

 

(a)

(b)