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What is the derivative of y = e^kx?
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Ch. 3 - Derivatives
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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TOC.tableOfContents
- 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: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 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?
Guida verificata passo dopo passo1
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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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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Percorso guidato
06:21
Properties of Functions
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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Percorso guidato
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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Percorso guidato
05:53Finding Differentials
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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.
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