Textbook Question
What is the derivative of y = e^kx?
Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 13c
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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Identify the relationship between the sides of an isosceles right triangle. The legs are equal, and the hypotenuse can be found using the Pythagorean theorem: \( c = \sqrt{2}a \), where \( a \) is the length of each leg and \( c \) is the hypotenuse.
Differentiate the Pythagorean theorem with respect to time \( t \) to find the rate of change of the hypotenuse. Start with \( c^2 = a^2 + a^2 = 2a^2 \).
Apply implicit differentiation to \( c^2 = 2a^2 \) with respect to \( t \): \( 2c \frac{dc}{dt} = 4a \frac{da}{dt} \).
Solve for \( \frac{dc}{dt} \), the rate of change of the hypotenuse: \( \frac{dc}{dt} = \frac{2a \frac{da}{dt}}{c} \).
Substitute the given rate of change of the legs \( \frac{da}{dt} = 2 \) m/s and the expression for \( c = \sqrt{2}a \) into the equation to find \( \frac{dc}{dt} \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Isosceles Right Triangle Properties
An isosceles right triangle has two equal sides and a right angle between them. The lengths of the legs are denoted as 'a', and the hypotenuse 'c' can be calculated using the Pythagorean theorem: c = a√2. Understanding these properties is essential for relating the sides of the triangle to each other.
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Related Rates
Related rates involve finding the rate at which one quantity changes in relation to another. In this problem, we need to determine how the length of the hypotenuse changes as the lengths of the legs increase. This requires applying differentiation to the relationship between the sides of the triangle.
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04:16Intro To Related Rates
Differentiation
Differentiation is a fundamental concept in calculus that deals with finding the rate of change of a function. In this context, we will differentiate the equation relating the legs and the hypotenuse with respect to time to find the rate at which the hypotenuse is changing as the legs grow.
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Related Practice
Textbook Question
Find f′(x) if f(x) = 15e^3x.
Textbook Question
The sides of a square decrease in length at a rate of 1 m/s.
b. At what rate are the lengths of the diagonals of the square changing?
Textbook Question
5–24. For each of the following composite functions, find an inner function u=g(x) and an outer function y=f(u) such that y=f(g(x)). Then calculate dy/dx.
y = (3x+7)¹⁰
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Textbook Question
7–14. Find the derivative the following ways:
a. Using the Product Rule (Exercises 7–10) or the Quotient Rule (Exercises 11–14). Simplify your result.
y = x² - 2ax +a² / x-a, where a is a constant
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Textbook Question
7–14. Find the derivative the following ways:
a. Using the Product Rule (Exercises 7–10) or the Quotient Rule (Exercises 11–14). Simplify your result.
y = x² - a² / x-a, where a is a constant