Ch. 3 - Derivatives

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

Hass 15th Edition

Ch. 3 - Derivatives

Problem 3.81

Chapter 3, Problem 3.81

For what value of c is the curve y = c/ (x + 1) tangent to the line through the points (0, 3) and (5, -2)?

Verified step by step guidance

First, find the equation of the line passing through the points (0, 3) and (5, -2). Use the slope formula: \( m = \frac{y_2 - y_1}{x_2 - x_1} \). Substitute the given points to find the slope.

With the slope \( m \) calculated, use the point-slope form of the line equation: \( y - y_1 = m(x - x_1) \). Substitute one of the points, say (0, 3), and the slope to find the equation of the line.

Next, find the derivative of the curve \( y = \frac{c}{x + 1} \) to determine its slope at any point \( x \). Use the quotient rule: \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} \), where \( u = c \) and \( v = x + 1 \).

Set the derivative of the curve equal to the slope of the line found in step 2. This will give you an equation involving \( c \) and \( x \). Solve for \( c \) in terms of \( x \).

Finally, find the point of tangency by equating the curve \( y = \frac{c}{x + 1} \) to the line equation found in step 2. Solve for \( x \) and substitute back to find the value of \( c \) that makes the curve tangent to the line.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Tangent Line

A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line at that point is equal to the derivative of the curve's function at that point. Understanding how to find the slope of the tangent line is crucial for determining the conditions under which the curve and the line are tangent.

Slope of a Line

The slope of a line is a measure of its steepness, calculated as the change in the y-coordinates divided by the change in the x-coordinates between two points on the line. For the line through the points (0, 3) and (5, -2), the slope can be calculated using the formula (y2 - y1) / (x2 - x1). This slope will be essential for comparing it to the slope of the tangent line to the curve.

Derivative

The derivative of a function at a point represents the rate of change of the function's value with respect to changes in its input. In this context, finding the derivative of the curve y = c/(x + 1) will allow us to determine the slope of the tangent line at any point on the curve. Setting this derivative equal to the slope of the line will help us find the value of c for which the curve is tangent to the line.

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Derivatives

Related Practice

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In Exercises 53 and 54, find dr/ds.

r cos 2s + sin²s = π

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Finding g on a small airless planet Explorers on a small airless planet used a spring gun to launch a ball bearing vertically upward from the surface at a launch velocity of 15 m/sec. Because the acceleration of gravity at the planet’s surface was gₛ m/sec², the explorers expected the ball bearing to reach a height of s = 15t − (1/2)gₛt² m t sec later. The ball bearing reached its maximum height 20 sec after being launched. What was the value of gₛ?

Textbook Question

Derivatives in Differential Form

In Exercises 17–28, find dy.

y = cos(x²)

Textbook Question

Approximation Error

In Exercises 29–34, each function f(x) changes value when x changes from x₀ to x₀ + dx. Find

a. the change Δf = f(x₀ + dx) − f(x₀);

b. the value of the estimate df = fʹ(x₀) dx; and

c. the approximation error |Δf − df|.

f(x) = x⁴, x₀ = 1, dx = 0.1

Textbook Question

In Exercises 41–58, find dy/dt.

y = (1/6)(1 + cos²(7t))³

Textbook Question

A melting ice layer A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 10 in³/min, how fast is the thickness of the ice decreasing when it is 2 in. thick? How fast is the outer surface area of ice decreasing?