Textbook Question
79–82. {Use of Tech} Visualizing tangent and normal lines
b. Graph the tangent and normal lines on the given graph.
x⁴ = 2x²+2y²; (x0, y0)=(2, 2) (kampyle of Eudoxus)
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 3.8.42b
Surface area of a cone The lateral surface area of a cone of radius r and height h (the surface area excluding the base) is A = πr√r²+h².
b. Evaluate this derivative when r=30 and h=40.
Verified step by step guidance1
First, identify the formula for the lateral surface area of a cone, which is given as A = πr√(r² + h²).
To find the derivative of A with respect to r, apply the chain rule. The expression inside the square root, r² + h², is a function of r, so differentiate it with respect to r.
The derivative of r² with respect to r is 2r, and since h is a constant, the derivative of h² with respect to r is 0. Therefore, the derivative of r² + h² with respect to r is 2r.
Now, differentiate the entire expression A = πr√(r² + h²) with respect to r. Use the product rule, which states that the derivative of a product u*v is u'v + uv'. Here, u = πr and v = √(r² + h²).
Evaluate the derivative at r = 30 and h = 40 by substituting these values into the derivative expression obtained in the previous step.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Lateral Surface Area of a Cone
The lateral surface area of a cone is the area of the cone's curved surface, excluding the base. It is calculated using the formula A = πr√(r² + h²), where r is the radius and h is the height of the cone. This formula derives from the geometry of the cone and involves the Pythagorean theorem to account for the slant height.
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Derivative
A derivative represents the rate of change of a function with respect to a variable. In this context, it helps determine how the lateral surface area A changes as the radius r varies, while keeping the height h constant. The derivative is calculated using differentiation rules, which provide insights into the behavior of the function at specific points.
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Evaluation of Derivatives
Evaluating a derivative involves substituting specific values into the derivative function to find the instantaneous rate of change at those points. In this case, after finding the derivative of the lateral surface area with respect to r, we will substitute r = 30 and h = 40 to compute the exact rate of change of the surface area at that radius.
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Related Practice
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Suppose f(3) = 1 and f′(3) = 4. Let g(x) = x2 + f(x) and h(x) = 3f(x).
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Textbook Question
13-26 Implicit differentiation Carry out the following steps.
b. Find the slope of the curve at the given point.
x = e^y; (2, ln 2)
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A bug is moving along the right side of the parabola y=x² at a rate such that its distance from the origin is increasing at 1 cm/min.
b. Use the equation y=x² to find an equation relating dy/dt to dx/dt.
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Textbook Question
{Use of Tech} Spring oscillations A spring hangs from the ceiling at equilibrium with a mass attached to its end. Suppose you pull downward on the mass and release it 10 inches below its equilibrium position with an upward push. The distance x (in inches) of the mass from its equilibrium position after t seconds is given by the function x(t) = 10sin t - 10cos t, where x is positive when the mass is above the equilibrium position. <IMAGE>
b. Find dx/dt and interpret the meaning of this derivative.
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