# F.4 A-MTH Differentiation (Quick)

1) If a hemispherical bowl of radius 10cm contains water to a depth of h cm, the volume of the water is (1/3))πh^2(30-h)cm^3. Water is poured into the bowl at a rate of 2cm^3/s.

(a) If the radius of the water surface is r cm, express r in terms of h.

(b) When the water depth is 4cm, find the rates of cnahge of (i) the water level(ii) the radius of the water surface.

2) Find the max and min values of y=2sin^2 x+ cos^2 x.

3) Given f(x) = x^3 + 3kx +5, prove that:

(a) If k>0, f(x) has no turning points.

(b) If k<0, f(x) has a max point and a min point.

Rating

(1)(a) Using the Pyth. thm:

r2 = 102 - (10 - h)2

r = √(20h - h2)

(b)(i) V = πh2(30 - h)/3

= π(30h2 - h3)/3

dV/dt = π(60h - 3h2)/3 x dh/dt

= π(20h - h2) dh/dt

When h = 4,

dV/dt = π(80 - 16) dh/dt

dh/dt = 1/(32π) cm/s

(ii) With r2 = 20h - h2, difftiating both sides give:

2r (dr/dt) = (20 - 2h) (dh/dt)

When h = 4, r = 8 and hence

16 (dr/dt) = 12/(32π)

dr/dt = 3/(128π) cm/s

(2) y = 2 sin2 x + cos2 x

= 1 + sin2 x

Therefore, max. of y = 2 and min. of y = 1 since 0 <= sin2 x <= 1.

(3)(a) f'(x) = 3x2 + 3k

If k > 0, f'(x) > 0 since x2 >= 0 for all real x.

So f'(x) = 0 has no real roots and therefore f has no turning points.

(b) f"(x) = 6x

f'(x) = 0

x2 = -k

x = √(-k) or -√(-k)

So f"(x) > 0 at x = √(-k) and f"(x) < 0 at x = -√(-k)

Hence, f(x) has a max. and a min. point.

Source(s): My Maths knowledge