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.
