Textbook Question
Use implicit differentiation to find dy/dx.
exy = 2y
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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 3.11.16
The edges of a cube increase at a rate of 2 cm/s. How fast is the volume changing when the length of each edge is 50 cm?
Verified step by step guidance1
First, identify the formula for the volume of a cube, which is V = a^3, where 'a' is the length of an edge.
Next, differentiate the volume formula with respect to time 't' to find the rate of change of the volume. This gives us dV/dt = 3a^2 * (da/dt).
Substitute the given rate of change of the edge length, da/dt = 2 cm/s, into the differentiated formula.
Substitute the given edge length, a = 50 cm, into the formula. This will allow you to calculate the rate at which the volume is changing at that specific edge length.
Finally, compute the expression to find the rate of change of the volume, dV/dt, using the substituted values.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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 volume of the cube changes as the length of its edges increases. This requires applying the chain rule of differentiation to relate the rates of change of the edge length and the volume.
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Intro To Related Rates
Volume of a Cube
The volume of a cube is calculated using the formula V = s^3, where s is the length of an edge. Understanding this formula is crucial because it allows us to express the volume in terms of the edge length, enabling us to differentiate it with respect to time to find the rate of change of volume.
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Differentiation
Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. In this context, we differentiate the volume function with respect to time to find how fast the volume is changing as the edge length changes. This process involves applying the power rule and the chain rule to relate the rates of change.
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Related Practice
Textbook Question
Calculate the derivative of the following functions (i) using the fact that bx = exIn b and (ii) using logarithmic differentiation. Verify that both answers are the same.
y = (x²+1)x
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Textbook Question
Match the graphs of the functions in a–d with the graphs of their derivatives in A–D. <MATCH A-D IMAGE>
241
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Textbook Question
Use Theorem 3.10 to evaluate the following limits.
lim x🠂0 (tan 5x) / x
335
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Textbook Question
Evaluate the derivative of the following functions.
f(x) = sec-1 (ln x)
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Textbook Question
Given that f(1) = 5, f′(1) = 4, g(1) = 2, and g′(1) = 3 , find d/dx (f(x)g(x))∣ ∣x=1 and d/dx (f(x) / g(x)) ∣ x=1.
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