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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 $100 cm^2?$

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

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Differentiation

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##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

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.

#### Topics

Derivatives

Differentiation

##### Catherine R.

Missouri State University

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp