Maths Sa . Msa1 Revision Applications Of Differentiation 1

Applications of Differentiation

SAJC MSA1 Revision (Mar 09)

Revision Questions: Applications of Differentiation 1

NJC 95\I\17(a) A hemispherical bowl is beneath a tap leaking at a constant rate. When the height of the water is h cm, the volume is π (rh 2 −

2

h3 ) cm 3, r being the radius of the 3

hemisphere. Find the rate at which the water level is rising when it is halfway to the top, given that r = 6 cm and that the bowl fills in 1 minutes. VJC 00/I/8 A television crew is televising a 100 m sprint event. xm S

F

10 m

θ C

3

The cameraman positions himself 10 m from F, a point on the finishing line. The point S represents the location of the first athlete to come in. He is running towards F such that angle SFC = . At time t s, SF = x m and angle FCS = θ . − When the athlete is 10 m away from F, he is running at 3π m s 1. Find the rate of 3π change of θ at this moment. { − } [4] 20 NYJC 95\I\9 An open rectangle box made of thin material has a square base, side x cm, and a capacity of 4000 cm 3. If A cm 2 is the area of the material used to make the box, show that A = x 2 +

4

16000 and find the value of x which makes the x

amount of material used a minimum. {x = 20} JJC 98/I/(b) A point C moves on the circumference of a emicircle of radius r in such a way that 0 0 < θ < 90 0 . Prove that the area of triangle ABC is equal to A

r2 [ 2 sin 2θ + sin 4θ ] . 4

5

Find θ when this area is maximum. AJC 97 P1 Q13(b) A circular hollow cone of height h cm and semi-vertical angle 30° stands on horizontal ground. A smaller upright cone with base radius x cm just fits into the cone as shown. (i) Show that the volume, V cm³ of the smaller cone is given by V

C



r

O

B

D [7]

30º

= 13 π x 2 (h − 3 x)

. (ii)

Prove that, as x varies, the maximum 4π 3 h possible volume of the smaller cone is 243

[7]

SAJC MSA1 Revision (Mar 09)

2

Applications of Differentiation

cm³. 2003AJC/P1/Q10(i),(ii)

x2 . 3x (i) Find the coordinates of the stationary points of the curve and determine their nature. {(0,0) is min pt ; (1.82, 0.449) is max pt} [4] 2 x (ii) Sketch the graph of y = x . [2] 3 4 2003CJC/P1/Q1 A petrol tanker is damaged in a road accident, and petrol leaks onto a flat section of a motorway. The leaking petrol begins to spread in a circle of thickness 0.002 m. Petrol is leaking from the tanker at a rate of 0.0084 m3s–1. Find the rate at which the radius of the circle of petrol is increasing at the instant when the radius of the circle is 3 m, giving your answer in ms–1 to 2 decimal places. [4] 21 2003SAJC/P1/Q9 Gas is being pumped into a spherical balloon at a constant rate of 15 m3/min. Calculate (i) the rate at which the surface area is increasing, at the instant when the radius is 2 m ; {15 m2/min} [4] (ii) the time taken for the radius to increase from 2 m to 4 m. {15.6 min} [3] A curve has equation y =

Applications of Differentiation

SAJC MSA1 Revision (Mar 09)

Solutions 6

h3 Volume of water V = π (6h − ) 3 dV dh = π (12 h − h 2 ) dt dt 2

2

3

π (6 ) 12 dV The bowl fills in 60s ⇒ = 3 = π dt 60 5 When the water level is halfway to the top, 12 dh π = π (12 (3) − 3 2 ) 5 dt dh 4 = dt 45 The water level is rising at

8

When x = 10 ,

4 cm per second. 45

dx = −3π dt

x 10 Differenti ating w.r.t. t , tan θ =

dθ 1 dx = dt 10 dt When x = 10 ,

sec 2 θ

(draw diagram! ) dθ 3π =− rad per second dt 20

9

A = Base +4 × Side = x 2 + 4 xh  4000  = x 2 + 4 x 2   x  16000 = x2 + (shown) x

To find the value of x that gives the minimum area, set the rate of change of the area, dA , to be 0 and solve for x. dx

dA 16000 = 2x − dx x2

At minimum,

dA =0 dx

2x −

16000 =0 x2 x 3 = 8000 x = 20

d2A at x = 20 dx 2 d2A 32000 =2+ = 6 > 0 (min. point) 2 dx x3

Check

Applications of Differentiation

SAJC MSA1 Revision (Mar 09)

10 JJC 98/I/(b) A point C moves on the circumference of a semicircle of radius r in such a way that 0 0 < θ < 90 0 . Prove that the area of triangle ABC is equal to

C

r2 [ 2 sin 2θ + sin 4θ ] . 4

A

Find θ when this area is maximum. ∠COD = 2 × ∠CAD = 2θ CB = sin ∠COD OC CB = OC sin ∠COD = r sin 2θ OB = cos ∠COD OC OB = OC cos ∠COD = r cos 2θ A=

r2 ( 2 sin 2θ + cos 4θ ) 4

Area =



r

O

B

D

1 × AB × CB 2

1 × ( AO + OB ) × CB 2 1 = × ( r + r cos 2θ ) × r sin 2θ 2 r2 = ( 2 sin 2θ + 2 sin 2θ cos 2θ ) 4 r2 = ( 2 sin 2θ + cos 4θ ) (shown) 4

=

In this case, please note that r is not the variable quantity. The variable quantity is only θ. d 2A r2 ( − 8 sin 2θ − 16 sin 4θ ) = 4 dθ 2

dA r 2 ( 4 cos 2θ + 4 cos 4θ ) = dθ 4

At the maximum,

dA r 2 ( 4 cos 2θ + 4 cos 4θ ) = 0 = dθ 4

cos 2θ + cos 4θ = 0 cos 2θ + 2 cos 2 2θ −1 = 0 2 cos 2 2θ + cos 2θ −1 = 0 ( 2 cos 2θ −1)( cos 2θ +1) = 0

cos 2θ = 0.5 2θ = 60 ° θ = 30 °

At θ = 30 ° ,

or

cos 2θ = −1 2θ =180 ° (invalid)

0°<θ<90° 0°<2θ<180°

d 2A r2 ( − 8 sin 60° − 16 sin 120 °) < 0 (maximum point) = 4 dθ 2