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 3.11.17
Textbook Question
A circle has an initial radius of 50 ft when the radius begins decreasing at a rate of 2 ft/min. What is the rate of change of the area at the instant the radius is 10 ft?
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1
Identify the formula for the area of a circle, which is given by A = πr², where A is the area and r is the radius.
Differentiate the area formula with respect to time t to find the rate of change of the area, dA/dt, using the chain rule: dA/dt = (dA/dr) * (dr/dt).
Calculate dA/dr, which is the derivative of the area with respect to the radius: dA/dr = 2πr.
Substitute the given rate of change of the radius, dr/dt = -2 ft/min (negative because the radius is decreasing), and the radius at the instant of interest, r = 10 ft, into the differentiated area formula.
Combine the values to find dA/dt at the moment when the radius is 10 ft, using the expression dA/dt = (2πr) * (dr/dt).
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