The area A of a circle is increasing at a constant rate of 2cm's-. If the area of the circle is given by A = Tr2, what is the rate of change of the radius when the radius is 4cm? %3D

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A: Given that, The spherical balloon is inflated and its rate of change of volume is; dVdt=30 (in3min)…

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A: Let V be the volume of the sphere which is given in the question as,

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A: Given,           The radius of a circle is increasing at the rate of 4 cm/s. And the radius is 5 cm…

Q: A circle has an initial radius of 50 ft when the radius begins decreasing at the rate of 4 ft/min.…

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Q: The radius of a circle is increasing at a constant rate of 8 centimeters per minute. At the instant…

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A: The given question can be solved as shown in step2.

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A: Area of circle of radius r inA=πr2→dAdr=2πrdrdt→2πrdrdt=6      [∵dAdr=6cm2/sec]    -(1)

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A: Area of a circle is given byDifferentiating the function of area with respect to time:

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A: As per guideline we have to solve one question Please repost other question again

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Q: The area of a square with sides of length s is given by A = s2 . Find the rate of change of the area…

A: The area of a square with sides of length s is given by , A=s2. To find the rate of change of area…

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Q: A circle has an initial radius of 50 ft when the radius begins decreasing at the rate of 5 ​ft/min.…

A: Given information:

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A: Given : V= 32π/3 cm3              dV/dt = 8 cm3 /sec

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A: We are given :  The Area of a square with  side s is A(s) =s2