A cylindrical can has a height of 19 cm and a radius of 6 cm. The volume of the cylindrical can is decreasing at a rate of 563 cubic cm per second, with the height being held constant. What is the rate of change of the radius, in cm per second, when the radius is 6 cm?

Round your answer to the nearest hundredth.

The rate of change of a function is simply the derivative of the function with respect to the changing parameter in the function i.e. variable in the function.

So for calculating the rate of change of radius, we calculate the rate of volume change as volume is the function of radius.

As we know volume of the cylinder is given by V=πr²h

where r is the radius and h is the height of the cyclinder.

Given that rate of change of volume = 563 cm/sec

⇒ dV/dt =563 cm/sec

⇒ d(πr²h)/dt =563

taking πh outside the derivative as π and h is constant term

⇒ πh dr²/dt =563

⇒ πh (dr²/dr)(dr/dt)= 563 (applying chain rule dy/dx = (dy/du)(du/dx))

⇒ πh(2r)(dr/dt) =563

⇒dr/dt= 563/(2πhr)

given height of the cylinder h= 19cm

rate of change of radius at radius r =6cm= dr/dt= 563/(2πhr)= 563/(2*π*19*6)= 0.79 cm/sec

Therefore rate of change of radius at radius 6cm is 0.79 cm/sec.

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