Rate of change of the angle

Question

Two sides of a triangle are 4 cm and 7 cm and the 3rd side is changing at 0.2cm/min. At what rate is the angle changing when the third side is 5 cm?

Use law of cosines

Answer 1

In this case, I'm assuming that you are studying derivatives now in the later part of your Precalculus.

The law of cosine is:

c² = a² + b² - 2ab cos (θ)

Given:

a=4 cm

b=7 cm

dc/dt = 0.2 cm/min

dθ/dt = ?

c² = 4² + 7² - 2(4)(7) cos (θ) = 16 + 49 - 56 cos (θ)

c² = 65 - 56 cos (θ)

If c=5 cm, then:

25 = 65 - 56 cos(θ)

25-65 = -56 cos(θ)

-40 = -56 cos (θ)

cos (θ) = -40/(-56) = 5/7

θ ≈ 44.42° ≈ 0.775 radians

For c² = 65 - 56 cos (θ), get the derivative of both sides of the equation with respect to time (t). We'll get:

2c•dc/dt = 56 sin (θ) • dθ/dt

substitute the value:

2(5)•(0.2) = 56 sin (0.775) • dθ/dt

2 = 56 sin (0.775) • dθ/dt

2 / (56 sin (0.775)) = dθ/dt

dθ/dt ≈ 0.051 radians / min. or 2.92° / min.

Note: Don't round off in your calculator until you reach the final answer.