Textbook Question
Volume of a torus The volume of a torus (doughnut or bagel) with an inner radius of a and an outer radius of b is V=π²(b+a)(b−a)²/4.
a. Find db/da for a torus with a volume of 64π².
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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 3.8.23a
13-26 Implicit differentiation Carry out the following steps.
a. Use implicit differentiation to find dy/dx.
³√x+³√y⁴ = 2;(1,1)
Verified step by step guidance1
Start by differentiating both sides of the equation with respect to x. The equation is \( \sqrt[3]{x} + \sqrt[3]{y^4} = 2 \). Remember that \( y \) is a function of \( x \), so when differentiating terms involving \( y \), use implicit differentiation.
Differentiate \( \sqrt[3]{x} \) with respect to \( x \). This is \( \frac{d}{dx}(x^{1/3}) \), which results in \( \frac{1}{3}x^{-2/3} \).
Differentiate \( \sqrt[3]{y^4} \) with respect to \( x \). This requires the chain rule: \( \frac{d}{dx}(y^4)^{1/3} = \frac{1}{3}(y^4)^{-2/3} \cdot 4y^3 \cdot \frac{dy}{dx} \).
Set the derivative of the left side equal to the derivative of the right side. Since the derivative of a constant (2) is 0, you have: \( \frac{1}{3}x^{-2/3} + \frac{4}{3}y^{3}(y^4)^{-2/3} \cdot \frac{dy}{dx} = 0 \).
Solve for \( \frac{dy}{dx} \) by isolating it on one side of the equation. This involves algebraic manipulation to express \( \frac{dy}{dx} \) in terms of \( x \) and \( y \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is a technique used to differentiate equations where the dependent and independent variables are not explicitly separated. Instead of solving for y in terms of x, we differentiate both sides of the equation with respect to x, treating y as a function of x. This method allows us to find the derivative dy/dx without isolating y.
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05:14
Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus that allows us to differentiate composite functions. When applying implicit differentiation, the chain rule is essential for differentiating terms involving y, as we must multiply by dy/dx when differentiating y with respect to x. This ensures that we account for the relationship between x and y in the differentiation process.
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05:02Intro to the Chain Rule
Evaluating Derivatives at a Point
After finding the derivative dy/dx using implicit differentiation, we often need to evaluate it at a specific point, such as (1,1) in this case. This involves substituting the x and y values into the derived expression for dy/dx to find the slope of the tangent line at that point. This step is crucial for understanding the behavior of the function at specific coordinates.
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04:50Critical Points
Related Practice
Textbook Question
79–82. {Use of Tech} Visualizing tangent and normal lines <IMAGE>
a. Determine an equation of the tangent line and the normal line at the given point (x0, y0) on the following curves. (See instructions for Exercises 73–78.)
x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)
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Textbook Question
A woman attached to a bungee cord jumps from a bridge that is 30 m above a river. Her height in meters above the river t seconds after the jump is y(t) = 15(1+e-t cos t), for t ≥ 0.
Determine her velocity at t = 1 and t = 3.
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Textbook Question
Derivatives and tangent lines
a. For the following functions and values of a, find f′(a).
f(x) = √3x; a= 12
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Textbook Question
Deriving trigonometric identities
a. Differentiate both sides of the identity cos 2t = cos² t−sin² t to prove that sin 2 t= 2 sin t cos t.
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Textbook Question
{Use of Tech} Tree growth Let b represent the base diameter of a conifer tree and let h represent the height of the tree, where b is measured in centimeters and h is measured in meters. Assume the height is related to the base diameter by the function h = 5.67+0.70b+0.0067b².
a. Graph the height function.
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