Q. 87 The聽area聽of聽a聽circle聽can聽b... [FREE SOLUTION]

Chapter 2: Q. 87 (page 211)

$The area of a circle can be written in terms of its radius as A = {蟺谤}^{2}, where both A and r are functions of time . Suppose a circular area from a spotlight on a stage floor is slowly expanding . (a) Find \frac{d A}{d r} and explain its meaning in practical terms . (b) Does the rate \frac{d A}{d r} depend on how fast the radius of the circle is increasing ? Does it depend on the radius of the circle ? Why or why not ? (c) Find \frac{d A}{d t} and explain its meaning in practical terms . (d) Does the rate \frac{d A}{d t} depend on how fast the radius of the circle is increasing ? Does it depend on the radius of the circle ? (e) If the radius of the circle of light is increasing at a constant rate of 2 inches per second, how fast is the area of the circle of light increasing at the moment that the spotlight has a radius of 24 inches ?$

Short Answer

Expert verified

$(a) . The value of \frac{d A}{d r} = 2 蟺谤 (b) . The rate \frac{d A}{d r} does not depends on how fast the radius of circle is increasing because rate of change of derivative of the area with respect to area and the rate \frac{d A}{d r} depends on the radius of circle because the value will increase or decrease as the radius changes . (c) . The value of \frac{d A}{d t} = 2 蟺谤 \frac{d r}{d t} (d) . The rate \frac{d A}{d t} depends on how fast the radius of circle is increasing because rate of change of derivative of the area with respect to area and the rate \frac{d A}{d t} depends on the radius of circle . (e) . The rate of change of area of circle of light = 96 蟺 .$

Step by step solution

Step 1. Given Information

$A = {蟺谤}^{2}$

Step 2. Solution (a) : value of dAdr

$Consider the equation : A = {蟺谤}^{2} Differentiate above equation \frac{d A}{d r} = \frac{d}{d r} ({蟺谤}^{2}) = 蟺 \frac{d}{dr} (r^{2}) = 2 蟺谤 inch The practical meaning is the rate of change of area of a circle with radius .$

Step 3. Solution (b) : How value of dAdr related to increase in radius of circle.

$The rate \frac{d A}{d r} does not depends on how fast the radius of circle is increasing because rate of change of derivative of the area with respect to area and the rate \frac{d A}{d r} depends on the radius of circle because the value will increase or decrease as the radius changes .$

Step 4. Solution (c) : Value of dAdt

$Consider the equation : A = {蟺谤}^{2} Differentiate above equation : \frac{d A}{d t} = \frac{d}{d t} ({蟺谤}^{2}) = 蟺 \frac{d}{d t} (r^{2}) = 2 蟺谤 \frac{d r}{d t} inch per seconds$

Step 5. solution (d) : How value of dAdtdepends on increase in radius

$The rate \frac{d A}{d t} depends on how fast the radius of circle is increasing because rate of change of derivative of the area with respect to area and the rate \frac{d A}{d t} depends on the radius of circle .$

Step 6. Solution (e) : The rate of change of area of circle of light.

$Consider the equation : A = {蟺谤}^{2} Differentiate above equation \frac{d A}{d t} = \frac{d}{d t} ({蟺谤}^{2}) = 蟺 \frac{d}{d t} (r^{2}) = 2 蟺谤 \frac{d r}{d t} inch per seconds Substitute the value in above equation {\frac{d A}{d t}}_{r = 24} = 2 蟺 (24) (2) = 96 蟺$

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91影视

Short Answer

Step by step solution

Step 1. Given Information

Step 2. Solution (a) : value of dAdr

Step 3. Solution (b) : How value of dAdr related to increase in radius of circle.

Step 4. Solution (c) : Value of dAdt

Step 5. solution (d) : How value of dAdtdepends on increase in radius

Step 6. Solution (e) : The rate of change of area of circle of light.

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