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Q19E

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Found in: Page 133

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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

The base of the triangle is changing at the rate of $$\frac{{db}}{{dt}} = - 1.6cm/min$$

See the step by step solution

## Step 1: Given Information.

It is given that increasing rate of altitude of triangle is $$1cm/min$$and that of the area is $$2cm/min$$. Rate change of base of triangle is to be found out.

## Step 2: Definition of Derivative

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value)..

## Step 3: Find derivative

Let $$h =$$altitude of the triangle

$$A =$$area of the triangle

$$b =$$area of the triangle

Now , area of the triangle

$$A = \frac{{bh}}{2}$$

Differentiate with respect to $$t$$

$$\begin{array}{l}\frac{d}{{dt}}\left( A \right) = \frac{d}{{dt}}\left( {\frac{{bh}}{2}} \right)\\\frac{{dA}}{{dt}} = \frac{{dh}}{{dt}}.\frac{b}{2} + \frac{h}{2}.\frac{{db}}{{dt}}\\\frac{h}{2}.\frac{{db}}{{dt}} = \frac{{dA}}{{dt}} - \frac{{dh}}{{dt}}.\frac{b}{2}\\\frac{{db}}{{dt}} = \frac{2}{h}.\frac{{dA}}{{dt}} - \frac{{dh}}{{dt}}.\frac{b}{h}\end{array}$$

Now,for the values

$$\begin{array}{l}h = 10cm\\A = 100c{m^2}\\b = \frac{{2.100}}{{10}}\\b = 20cm\end{array}$$

$$\begin{array}{c}\frac{{db}}{{dt}} = \frac{2}{h}.\frac{{dA}}{{dt}} - \frac{{dh}}{{dt}}.\frac{b}{h}\\ = \frac{2}{{10}}.2 - 1.\frac{{20}}{{10}}\\ = - \frac{8}{5}\\ = - 1.6\end{array}$$

Therefore, the base of the triangle is changing at the rate of $$\frac{{db}}{{dt}} = - 1.6cm/min$$