A lvl H2 Maths: Differentiation

TJC C Maths Prelims 1998

(a) Find the equation of the normal to the curve cos (x²) + xy² = 2y at the point where the curve cuts the y-axis.

(b) A circular cylinder is inscribed in a sphere with radius 26 cm so that all points on the circumference of the two circular ends are on the surface of the sphere. At a certain instant, the radius of the cylinder is 24 cm and is decreasing at the rate of 0.5 cm s^-1.
(i) Find the rate at which the height is changing at that instant.
(ii) Find the exact value of the radius of the cylinder when its curved surface area is a maximum.

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Answer:

(a) When the curve cuts the y-axis, x = 0
Hence, cos 0 + 0 = 2y
1 = 2y
y = ½

Thus, curve cuts y-axis at (0, ½)

cos (x²) + xy² = 2y
Differentiate w.r.t. x
-sin (x²) * 2x + x (2y dy/dx) + y² = 2 dy/dx
y² - 2x sin (x²) = (dy/dx) (2 - 2xy)
$\frac{dy}{dx} = \frac{y^{2}-2x sin\left(x^{2} \right)}{2 - 2xy}$

When x = 0, y = ½, dy/dx = ⅛,
gradient of normal = -8

Hence, equation of normal is
y - ½ = -8 (x - 0)
y = -8x + ½

(b)
(i) Let x = radius, h = height
Given dy/dx = -0.5

$h = \sqrt{52^{2} - 4x^{2}}$
$\frac{dh}{dx} = \frac{1}{2}\left(52^{2} - 4x^{2} \right)^{-\frac{1}{2}}\left(-8x \right)$
$\frac{dh}{dx} = -\frac{4x}{\sqrt{52^{2} - 4x^{2}}} = \frac{-4x}{h}$

When x = 24, dh/dx = -48
dh/dt = dh/dx * dx/dt
= 2.4 cm/s

Thus, the height is increasing at a rate of 2.4 cms^-1.

(ii) Let A = curved surface area.

A = 2πxh
$\frac{dA}{dx} = 2\pi x \left(\frac{dh}{dx} \right) + h\left(2\pi \right)\left(1 \right)$
$= \frac{-8\pi x^{2}}{h} + 2\pi h$
-4x² + 2h² = 0
4x² = h²
4x² = 52² - 4x²
8x² = 52²
x² = 338
x = 13√2