Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.7.43

Chapter 3, Problem 3.7.43

The eight curve Find the slopes of the curve y⁴ = y² – x² at the two points shown here.

Verified step by step guidance

To find the slope of the curve at a given point, we need to find the derivative of the curve equation with respect to x. The given equation is y⁴ = y² - x².

Differentiate both sides of the equation with respect to x. Use implicit differentiation: d/dx(y⁴) = d/dx(y² - x²).

Apply the chain rule to differentiate y⁴ and y²: 4y³(dy/dx) = 2y(dy/dx) - 2x.

Rearrange the equation to solve for dy/dx: dy/dx = (2x) / (4y³ - 2y).

Substitute the coordinates of the given points into the derivative to find the slope at each point. For the point (√3/4, √3/2), substitute x = √3/4 and y = √3/2. For the point (√3/4, 1/2), substitute x = √3/4 and y = 1/2.

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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 isolated. In this case, the equation y⁴ = y² - x² involves both x and y, making it necessary to apply implicit differentiation to find the slope of the curve at specific points. This method allows us to differentiate both sides of the equation with respect to x, treating y as a function of x.

Slope of a Curve

The slope of a curve at a given point is defined as the derivative of the function at that point. It represents the rate of change of y with respect to x. For the curve defined by the equation y⁴ = y² - x², finding the slope at the specified points involves evaluating the derivative obtained from implicit differentiation and substituting the coordinates of the points into this derivative.

Critical Points

Critical points are values of x where the derivative of a function is either zero or undefined. In the context of the given curve, identifying critical points can help determine where the slope changes, which is essential for understanding the behavior of the curve. The points provided in the question are critical for evaluating the slope, as they represent specific locations on the curve where the slope needs to be calculated.

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Critical Points

Related Practice

Textbook Question

For what value or values of the constant m, if any, is

ƒ(x) = { sin 2x, x ≤ 0

{ mx, x > 0

a. continuous at x = 0?

b. differentiable at x = 0?

Give reasons for your answers.

Textbook Question

Derivatives in Differential Form

In Exercises 17–28, find dy.

y = x√(1 − x²)

Textbook Question

Differential Estimates of Change

In Exercises 35–40, write a differential formula that estimates the given change in volume or surface area.

The change in the lateral surface area S = 2πrh of a right circular cylinder when the height changes from h₀ to h₀ + dh and the radius does not change

Textbook Question

Second Derivatives

In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.

3 + sin y = y – x³

Textbook Question

Free-Fall Applications

Free fall on Mars and Jupiter The equations for free fall at the surfaces of Mars and Jupiter (s in meters, t in seconds) are s = 1.86t² on Mars and s = 11.44t² on Jupiter. How long does it take a rock falling from rest to reach a velocity of 27.8 m/sec (about 100 km/h) on each planet?

Textbook Question

Derivatives

In Exercises 23–26, find dr/dθ.

r = (1 + sec θ) sin θ