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The altitude of a triangle is increasing at a rate of $ 1 cm/min $ while the area of the triangle is increasing at a rate of $ 2 cm^2/min. $ At what rate is the base of the triangle changing when the altitude is $ 10 cm $ and the area is $ 100 cm^2? $

$-1.6 \mathrm{cm} / \mathrm{min}$

00:53

Amrita B.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 9

Related Rates

Derivatives

Differentiation

Campbell University

Baylor University

University of Michigan - Ann Arbor

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in order to figure out the rate. The first thing we know is that h is altitude and abuse base. Looking at the formula 1/2 base times height, we can plug in, let me know 100 centimeters squared his area. I only know H is 10 centimetres. Giving us B is 20 centimeters. Therefore, into the equation D A over DT is 1/2 HDB over DT. So it's 10 times D B over DT plus B times D H over DT 20 times one was just 20. Simplify this. We got a DP over. DT is negative 1.6 centimeters her minutes. So clearly the rate is decreasing since native.

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