APPLICATION OF DERIVATIVES
(SUM 1 TO 4)
ALL EDUCATION BOARD
Question 1 Find the rate of change of the area of a circle with respect to its radius r. How Fast is the Area changing with respect to the radius when the radius is 3 cm?
Solution :
Radius of circle =` r`
Area of circle = A =` \Pi r^{2} `
Rate of change of area w.r.t.r=` \frac{dA}{dr} = \frac{d}{dr} ( \Pi r^{2} )=2 \Pi r`
Rate of change of area when r is `3 cm =2 \Pi \times 3 = 6 \Pi cm^{2} `/cm
Question 2 Find the rate of change of volume of the ball with respect to its radius r.How fast is the volume changing with respect to the radius when tyhe radius is 2 m ?
Solution :
Radius of the ball =r
Volume of the ball,V = `\frac{4}{3} \Pi r^{3} `
` Rate of change of volume w.r.t.r= ` \frac{dV}{dr}= \frac{d}{dr}(\frac{4}{3} \Pi r^{3})`
=`\frac{4}{3} \Pi .3 r^{2} = 4 \Pi r^{2} `
Rate of change of volume when r is 2m = `4 \Pi (2^{2}) =16 \Pi m^{3}` /m
Question 3 A balloon which always remains spherical has a variable radius ,Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Solution :
Let r be the radius V be the volume of the ballooon.
`V=\frac{4}{3} \Pi r^{3}`
Rate of change (increasing) of volume w.r.t.r
=`\frac{dV}{dr} = \frac{d}{dr} (\frac{4}{3} \Pi r^{3} )= 4 \Pi r^{2}`
Rate of change of volume when r is 10 cm = `4 \Pi (10)^{2} =400 \Pi cm^{3}` / cm
Question 4. The radius of a balloon is increasing at the rate of 10 cm per second. At what rate is the surface area of the balloon increasing when its radius is 15 cm.
solution :
Solution. Let be the radius of the balloon and S be the surface area of balloon at time t
`S= 4 \Pi r^{2} `
Rate of change (increase) of the radius w.r.t. . t = 10 cm/sec. (Given)
`\frac{dr}{dt} =10`
Rate of change of surface area w.r.t.
t `= \frac{dS}{dt} =8 \Pi r . \frac{dr}{dt} `
`=8 \Pi r \times 10 = 80 \Pi r`
Rate of change of surface area when ris 15 cm
`=80 \Pi (15) cm^{2}` /sec
`=1200 \Pi cm^{2}` /sec
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