Question

The base b(t)b(t)b, (, t, )of a triangle is decreasing at a rate of 131313 millimeters per minute and the height h(t)h(t)h, (, t, )of the triangle is increasing at a rate of 666 millimeters per minute. At a certain instant t_0t 0 ​ t, start subscript, 0, end subscript, the base is 555 millimeters and the height is 111 millimeter. What is the rate of change of the area A(t)A(t)A, (, t, )of the triangle at that instant (in square millimeters per minute)?

Answer 1

Answer:

  8.5 square mm per minute

Step-by-step explanation:

A = 1/2bh

A' = (1/2)(b'h +bh') . . . . differentiating with respect to time

A' = (1/2)((-13)(1) +(5)(6)) = (1/2)(17) = 8.5 . . . . mm²/min

The area is increasing at the rate of 8.5 mm²/min.

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Numerical solution

Consider the triangle dimensions .01 minutes before and after the time of interest.

  before: base = 5 + 0.13 = 5.13 mm; height = 1 - 0.06 = 0.94 mm. area = 2.4111 mm²

  after: base = 5 -0.13 = 4.87 mm; height = 1 +.06 = 1.06 mm. area = 2.5811 mm²

Average rate of change in that period is

  (2.5811 -2.4111)/0.02 = .17/.02 = 8.5 . . . . mm²/min

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Comment on the numbers

Note that we have assumed the base is decreasing at 13 mm/min. Quite often, division bars get lost on Brainly, so we're not sure this isn't 1/3 mm/min. At the assumed rate, the triangle will disappear in 5/13 minutes, about 23.08 seconds, when the base becomes zero. Of course, it only came into existence 1/6 minute (10 seconds) ago, when the height became non-zero.