The radius of a circle is increasing uniformly at the rate of 3
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A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 Cm, how fast is the enclosed area increasing?

Let r be the radius of the circular wave and A the enclosed area at time t.
 therefore space space space space space space straight A space equals space πr squared space space space space space space space space space space space space space space rightwards double arrow space space space space space space space space space dA over dt space equals space straight pi space. space 2 straight r space dr over dt space equals space 2 space πr space dr over dt
From given condition,    dr over dt space equals space 5 space cm divided by sec
therefore space space space space space space space when space straight r space equals space 8 space cm comma space space space dA over dt space equals space 2 straight pi cross times 8 cross times 5 space cm squared divided by straight s space equals space 80 space cm squared divided by straight s
therefore space space space space the space enclosed space area space is space increasing space at space the space rate space of space 80 space cm squared divided by straight s comma space space when space straight r space equals space 8 space cm.

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The volume of a cube is increasing at the rate of 8 cm3/s. How fast is the surface area increasing when the length of an edge is 12 cm?

Let V be volume of cube of side x.
therefore space space space space space space space space space space space space space straight V space equals space straight x cubed
From given condition,
                dV over dt space equals space 8 space cm cubed divided by straight s
therefore space space space space straight d over dt left parenthesis straight x cubed right parenthesis space equals space 8 space space space space space space space space space space space space rightwards double arrow space space space space 3 straight x squared dx over dt space equals space 8 space space space rightwards double arrow space space space space dx over dt space equals fraction numerator 8 over denominator 3 straight x squared end fraction space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Let S be surface area of cube
therefore space space space space space straight S space equals space 6 straight x squared
Rate of increase of surface area  = dS over dt space equals space straight d over dt left parenthesis 6 straight x squared right parenthesis
                                             equals space 12 straight x space dx over dt space equals space 12 straight x space cross times space fraction numerator 8 over denominator 3 straight x squared end fraction space space space space space space open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets
equals space 32 over straight x
When space straight x space equals space 12 comma space space rate space of space increase space of space surface space area space space equals space 32 over 12 equals space 8 over 3 cm squared divided by straight s.

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Since the rate of change is represented by derivative with respect to time, it is taken to be positive if the quantity is increasing and negative if the quantity is decreasing.
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An edge of a variable cube is increasing at the rate of 3 cm per second. How fast is the volume of the cube increasing when edge is 10 cm long?


Let x cm. be the length of an edge of a variable cube and V be its volume at any time t.
therefore space space space space space straight V space equals space straight x cubed space space space space space space space space space space space space space space space rightwards double arrow space space space space space dV over dt space equals space 3 straight x squared dx over dt
But   dx over dx space equals space 3 space cm divided by straight s                                                       (given)

therefore space space space space space dV over dt space equals space 3 straight x squared.3 space equals space 9 straight x squared
When x = 10 cm,   dV over dt space equals space 9 cross times left parenthesis 10 right parenthesis squared space equals space 900 space cm cubed divided by straight s
∴   volume of the cube is increasing at the rate of 900 cm3/s when the edge is 10 cm long.
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The radius of a circle is increasing uniformly at the rate of 3 cm per second. Find the rate at which the area of the circle is increasing when the radius is 10 cm.


Let r cm be the radius of the circle.
From given condition, rate of increase  = 3 cm per second
therefore space space space space space space dr over dt space equals space 3
Let A be the area of circle,               therefore space space space space straight A space equals space πr squared
therefore space space space space space rate space of space increase space of space area space space equals space dA over dt space equals space straight d over dt left parenthesis πr squared right parenthesis space equals space 2 space πr space dr over dt
space space space space space space space space
                            equals space 2 πr cross times 3 space equals space 6 πr                                     open square brackets because space space of space left parenthesis 1 right parenthesis close square brackets
When r = 10,   rate of increase of area  = 6 straight pi cross times 10 space equals space 60 straight pi space cm squared divided by sec
 

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A balloon which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the latter is 10 cm.


Let V be volume of balloon of radius r
therefore space space space space space space straight V space equals space 4 over 3 πr cubed
Rate of increase of volume w.r.t radius
 equals space dV over dr space equals straight d over dr open parentheses 4 over 3 πr cubed close parentheses
space equals space fraction numerator 4 straight pi over denominator 3 end fraction cross times space 3 straight r squared space equals space 4 πr squared
when r = 10 cm, rate of increase of volume = 4 space straight pi space left parenthesis 10 right parenthesis squared space equals space 400 space straight pi space cm cubed divided by cm

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