O lvl A Maths: Applications of Differentiation (Rates of Change)

A man 1.5m tall is walking at a speed of 2 ms^-1 away from a lamppost which has a lamp 5m above the ground, as shown in the figure.

Find
(a) the rate at which his shadow is increasing,
(b) the speed of the top of his shadow.

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Answer:

Let the length of shadow = x
Let the distance from the lamppost to the man = y

Since the man is 1.5 m tall,
using similar triangles,

(x+y)/5 = x/1.5
y/5 = 7x/15
y = 7x/3

(a) Given: dy/dt = 2
To find: dx/dt

From y=7x/3, dy/dx = 7/3

dy/dt = dy/dx * dx/dt
2 = 7/3 * dx/dt
dx/dt = 6/7 ms^-1

Thus, the shadow is increasing at 6/7 ms^-1


(b) Let z = distance of top of shadow from lamppost
z = x + y

Thus,
speed of the top of his shadow
= dz/dt
= d(x + y) / dt
= dx/dt + dy/dt
= 20/7 ms^-1

O lvl A Maths: Applications of Differentiation (Rates of Change)

The figure shows the curve y² = 2x. The points O (0,0) and P(x, y) lie on the curve. Given that the coordinates of Q is (x, 0),

(a) express the area, A cm², of the triangle OPQ in terms of x.
(b) If x is increasing at a rate of 4 units per seconds, calculate the rate of increase of A when x = 2.

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Answer:

(a) Area of triangle OPQ, A = ½(OQ)(PQ) = ½xy
y² = 2x


Since area of triangle is positive, we take the positive value of y

Hence,



(b) Using product rule,





If x is increasing at a rate of 4 units per seconds, then dx/dt = 4.
(chain rule)



When x = 2, dA/dt = 6(2) / √(2)(2)
= 6

O lvl A Maths: Applications of Differentiation (Rates of Change)

The lengths of the base and height of a right-angled triangle are (x - 1) cm and (x + 2) cm respectively. The length x cm at time t seconds is given by the expression x = 4t + 1, where t ≥ 0.

(a) Show that the area of the triangle in terms of t is given by A = 2t (4t + 3)
(b) Hence, find the rate of change of the area at the instant when t = 2.

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Answer:

(a) Area of triangle, A = ½(x - 1)(x + 2)

Substituting x = 4t + 1 into the equation
A = ½(4t + 1 - 1)(4t + 1 + 2)
= ½ (4t)(4t + 3)
= 8t² + 6t
= 2t (4t + 3)
(shown)


(b) A =8t² + 6t
dA/dt = 16t + 6

When t = 2, dA/dt = 16(2) + 6 = 38

Thus, the area of the triangle is increasing at a rate of 38 cm² per second at the instant when t = 2.