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 13c
Textbook Question
The legs of an isosceles right triangle increase in length at a rate of 2 m/s.
c. At what rate is the length of the hypotenuse changing?
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1
Identify the relationship between the legs of the isosceles right triangle and the hypotenuse using the Pythagorean theorem: if the legs are of length 'x', then the hypotenuse 'h' can be expressed as h = x√2.
Differentiate both sides of the equation h = x√2 with respect to time 't' to find the rate of change of the hypotenuse: dh/dt = (d/dt)(x√2).
Apply the chain rule to differentiate the right side: dh/dt = √2 * (dx/dt), where dx/dt is the rate of change of the leg length.
Substitute the known rate of change of the leg length (dx/dt = 2 m/s) into the differentiated equation to find dh/dt.
Calculate the value of dh/dt to determine the rate at which the hypotenuse is changing.
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