Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals3h 25m
4. Applications of Derivatives
Related Rates
Problem 112
Textbook Question
A spherical balloon is inflated at a rate of 10 cm³/min. At what rate is the diameter of the balloon increasing when the balloon has a diameter of 5 cm?
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1
Identify the relationship between the volume of the balloon and its diameter. The volume V of a sphere is given by the formula V = (4/3)πr³, where r is the radius.
Since the diameter d is related to the radius by the equation d = 2r, express the radius in terms of the diameter: r = d/2.
Differentiate the volume formula with respect to time t using the chain rule. This will involve differentiating V with respect to r and then r with respect to d.
Substitute the known rate of change of volume (dV/dt = 10 cm³/min) and the diameter (d = 5 cm) into the differentiated equation to find the rate of change of the radius (dr/dt).
Convert the rate of change of the radius (dr/dt) to the rate of change of the diameter (dd/dt) using the relationship dd/dt = 2(dr/dt).
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