Two sides of a triangle are #6m# and #7m# in length and the angle between them is increasing at a rate of #0.06# rad/s. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is #π/3# rad?

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Answer 1 · 2017-03-24T03:06:22

Area is increasing at rate of #0.63 \ m^2s^(-1)# (4dp)

Let us set up the following variables:

# {(theta, "Angle between the two known sides","radians"), (A, "Area of the triangle",m^2), (t, "time",sec) :} #

The area of the triangle is:

# A = 1/2(6)(7) sin theta = 21sin theta #

Differentiating wrt #theta# we get:

# (dA)/(d theta) = 21cos theta #

# (dA)/dt = (dA)/(d theta) * (d theta)/dt #

# :. (dA)/dt = 21cos theta * 0.06 #
# " " = 1.26cos theta #

So when #theta = pi/3 => #

# (dA)/dt = 1.26cos (pi/3) #
# " " = 1.26 * 1/2 #
# " " = 0.63 #