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rewona [7]
3 years ago
12

Approximate the change in the volume of a sphere when its radius changes from r​ = 40 ft to r equals 40.05 ft (Upper V (r )equal

s four thirds pi r cubed ). When r changes from 40 ft to 40.05 ​ft, Upper DeltaValmost equals nothing ftcubed.
Mathematics
1 answer:
alexgriva [62]3 years ago
7 0

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere (V), measured in cubic feet, is represented by the following formula:

V = \frac{4\pi}{3}\cdot r^{3}

Where r is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

\Delta V = \frac{\partial V}{\partial r}\Delta r

The derivative of the volume as a function of radius is:

\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}

Then, the change in volume is expanded:

\Delta V = 4\pi \cdot r^{2}\cdot \Delta r

If r = 40\,ft and \Delta r = 40\,ft-40.05\,ft = 0.05\,ft, the change in the volume of the sphere is approximately:

\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)

\Delta V \approx 1005.310\,ft^{3}

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

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