Differential Estimates of Change

In Exercises 35–40, write a differential formula that estimates the given change in volume or surface area.

The change in the volume V = (4/3)πr³ of a sphere when the radius changes from r₀ to r₀ + dr

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Start by recalling the formula for the volume of a sphere: \( V = \frac{4}{3} \pi r^3 \).

To find the differential \( dV \), differentiate the volume formula with respect to \( r \). This gives \( dV = \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) \cdot dr \).

Calculate the derivative: \( \frac{d}{dr} \left( \frac{4}{3} \pi r^3 \right) = 4 \pi r^2 \).

Substitute the derivative back into the differential formula: \( dV = 4 \pi r^2 \cdot dr \).

This differential formula \( dV = 4 \pi r^2 \cdot dr \) estimates the change in volume when the radius changes from \( r_0 \) to \( r_0 + dr \).

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Differential calculus focuses on the concept of the derivative, which represents the rate of change of a function. In this context, it helps estimate how a small change in one variable, such as the radius of a sphere, affects another variable, like the volume. Understanding derivatives is crucial for applying differential formulas to approximate changes.

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The volume of a sphere is given by the formula V = (4/3)πr³, where r is the radius. This formula is essential for understanding how the volume changes with respect to the radius. Knowing this relationship allows us to apply calculus concepts to find the differential change in volume when the radius changes slightly.

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