Thank you! The altitude of a triangle is increasing at a rate of 3.000 centimete

Question

Thank you!

The altitude of a triangle is increasing at a rate of 3.000 centimeters/minute while the area of the triangle is increasing at a rate of 1.000 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 9.000 centimeters and the area is 94.000 square centimeters? Note: The "altitude" is the "height" of the triangle in the formula "Area=(l/2)*base*height". Draw yourself a general "representative" triangle and label the base one variable and the altitude (height) another variable. Note that to solve this problem you don't need to know how big nor what shape the triangle really is.

Explanation / Answer

Let: Altitude of the triangle = h Base of the triangle = b Area of the triangle = A Now:                         Areaof the triangle = A = (1/2)bh ........(1) Differentiating A with respect to time:                            dA/dt= (1/2)[b(dh/dt)+h(db/dt)]......... (2) It is given that: dh/dt = 2cm / min dA/dt = 2 cm2 / min h = 10 cm A = 90 cm2 For A = 90 cm2 and h = 10 cm, we get b from(1):                           b = 2A/h = 2 (90) / 10 = 18 cm Now put these values in (2):                           2 = (1/2) [(18)(2) + (10)(db/dt)]                            4= 36 + 10 (db/dt) this implies     db/dt = -32/10 =-3.2 cm / min So the length of the base is decreasing by 3.2 cm permin. For A = 90 cm2 and h = 10 cm, we get b from(1):                           b = 2A/h = 2 (90) / 10 = 18 cm Now put these values in (2):                           2 = (1/2) [(18)(2) + (10)(db/dt)]                            4= 36 + 10 (db/dt) this implies     db/dt = -32/10 =-3.2 cm / min So the length of the base is decreasing by 3.2 cm permin.

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