# Thread: Difficult rates of change

1. ## Difficult rates of change

1. If you know two sides, a and b, of a triangle, and the angle, X, between them, you can calculate the area, A, by the formula A = (1/2)ab sin X. Suppose that at 2pm one of the sides is 6cm and is increasing at 2cm per second, another side is 10 cm and is increasing at 1cm per second, and the area is 15cm squared and is decreasing at 3cm squared per second. What is the angle between the two sides at 2pm, and at what rate is it changing?

2. At noon, a cylinder of variable dimensions has a radius of 13cm and a height of 12cm. The radius is increasing at a rate of 1cm per second, and the height is decreasing at the rate of 2cm per second. At what rate is the volume changing?

Question 2 has me stumped I made an attempt but it didnt work out.

Question 1 I just cant make sense of but I figure X is either pi/6 or 5pi/6

Thanks for everyone who helps in advance

2. These are both related rates questions. For Q1, A(t)=1/2a(t)b(t)sinX(t). Differentiate wrt to t. Q2 can be solved with a similar approach.

3. Hello, MC Squidge!

2. At noon, a cylinder has a radius of 13cm and a height of 12 cm.
The radius is increasing at a rate of 1 cm/sec,
and the height is decreasing at the rate of 2 cm/sec.
At what rate is the volume changing?

We are given: .r = 13, . dr/dt = +1

. . . . . . . . . . .h = 12, . dh/dt = -2

The volume of a cylinder is: .V .= .π(r^2)(h)

Then: . dV/dt .= .2π(r)(h)(dr/dt) + π(r^2)(dh/dt)

Hence: . dV/dt .= .2π(13)(12)(+1) + π(13^2)(-1) .= .312π - 338π

Therefore: .dV/dt .= .-26π cm^3/sec