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 3.8.19b
13-26 Implicit differentiation Carry out the following steps.
b. Find the slope of the curve at the given point.
cos y = x; (0, π/2)
Verified step by step guidance1
Start by differentiating both sides of the equation with respect to x. For the left side, differentiate cos(y) using the chain rule: the derivative of cos(y) is -sin(y) times the derivative of y with respect to x, which is dy/dx. For the right side, the derivative of x is 1.
Write the differentiated equation: -sin(y) * (dy/dx) = 1.
Solve for dy/dx by isolating it on one side of the equation. This involves dividing both sides by -sin(y), giving dy/dx = -1/sin(y).
Substitute the given point (0, π/2) into the equation to find the slope at that point. Since y = π/2, substitute π/2 for y in the expression for dy/dx.
Evaluate sin(π/2) to find the slope. Since sin(π/2) = 1, substitute this value into the expression for dy/dx to find the slope at the point (0, π/2).
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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 one variable in terms of the other, we differentiate both sides of the equation with respect to the independent variable, applying the chain rule as necessary. This method is particularly useful for curves defined by equations that cannot be easily rearranged.
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Finding The Implicit Derivative
Slope of a Curve
The slope of a curve at a given point represents the rate of change of the dependent variable with respect to the independent variable at that specific point. Mathematically, it is found by evaluating the derivative of the function at that point. In the context of implicit differentiation, the slope can be expressed as dy/dx, which indicates how y changes with respect to x.
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11:41Summary of Curve Sketching
Evaluating Derivatives at a Point
To find the slope of a curve at a specific point, we first compute the derivative of the function and then substitute the coordinates of the point into this derivative. This process allows us to determine the instantaneous rate of change at that point. For the given problem, substituting the point (0, π/2) into the derived expression for dy/dx will yield the slope of the curve at that location.
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Related Practice
Textbook Question
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.
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Textbook Question
Suppose f(3) = 1 and f′(3) = 4. Let g(x) = x2 + f(x) and h(x) = 3f(x).
Find an equation of the line tangent to y = h(x) at x = 3.
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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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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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Textbook Question
21–30. Derivatives
b. Evaluate f'(a) for the given values of a.
f(x) = 1/x+1; a = -1/2;5