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A cylindricial can has a height of 18cm and a radius of 4cm. The colume of the cylindrical can is decreasing at a rate of 560 cubic cm per seond, with the height being held constant. What is the reate

Answer :

Final answer:

This answer explains the process to find the rate at which the radius of a cylindrical can is decreasing, given its volume reduction rate, using the principles of calculus specifically related to related rates.

Explanation:

The question involves finding the rate at which the radius of a cylindrical can is decreasing given that its volume is decreasing at a rate of 560 cubic cm per second while the height remains constant at 18cm. To start, recall the formula for the volume of a cylinder, V = πr²h. Since the volume is changing, we identify this as a related rates problem in calculus, requiring differentiation with respect to time (t). Differentiating both sides of the volume formula with respect to time, we obtain dV/dt = 2πrh(dr/dt) + πr²(dh/dt). Given that the height (h) is constant, dh/dt = 0, simplifying our equation to dV/dt = 2πrh(dr/dt). Then, substituting the given values (dV/dt = -560 cm³/s, h = 18 cm, and r = 4 cm), we solve for dr/dt to find the rate at which the radius is decreasing.

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