Q. 23 In Exercises 19鈥�26, write down... [FREE SOLUTION]

Chapter 3: Q. 23 (page 299)

In Exercises 19鈥�26, write down an equation that relates the two quantities described. Then use implicit differentiation to obtain a relationship between the rates at which the quantities change over time. The surface area S and radius r of a cone with a fixed height of 5 units

Short Answer

Expert verified

Equation relating the quantities: $S = 蟺谤 \sqrt{r^{2} + 25} + {蟺谤}^{2}$

Implicit differentiation: $\frac{d S}{d t} = (蟺 \sqrt{r^{2} + 25} + \frac{{蟺谤}^{2}}{\sqrt{r^{2} + 25}} + 2 蟺 r) \frac{d r}{d t}$

Step by step solution

Step 1. Given information

Surface area of the cone $= S$

Radius of the cone $= r$

Height of the cone $= 5$

Step 2. Equation relating the quantities

Surface area of the cone is given by,

$S = 蟺谤 \sqrt{r^{2} + h^{2}} + {蟺谤}^{2} 鈬� S = 蟺谤 \sqrt{r^{2} + 5^{2}} + {蟺谤}^{2} 鈬� S = 蟺谤 \sqrt{r^{2} + 25} + {蟺谤}^{2}$

Step 3. Implicit differentiation with respect to time

From the above step,

$S = 蟺谤 \sqrt{r^{2} + 25} + {蟺谤}^{2} Differentiating on both sides, we get, \frac{d S}{d t} = 蟺 \sqrt{r^{2} + 25} \frac{d r}{d t} + \frac{蟺谤 (2 r)}{2 \sqrt{r^{2} + 25}} \frac{d r}{d t} + (2 r) 蟺 \frac{d r}{d t} 鈬� \frac{d S}{d t} = (蟺 \sqrt{r^{2} + 25} + \frac{{蟺谤}^{2}}{\sqrt{r^{2} + 25}} + 2 蟺 r) \frac{d r}{d t}$

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In Exercises 19鈥�26, write down an equation that relates the two quantities described. Then use implicit differentiation to obtain a relationship between the rates at which the quantities change over time. The surface area S and radius r of a cone with a fixed height of 5 units

Short Answer

Step by step solution

Step 1. Given information

Step 2. Equation relating the quantities

Step 3. Implicit differentiation with respect to time

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