Textbook Question
Find the derivatives of the functions in Exercises 1–42.
______
𝓻 = √2θ sinθ
278
views
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
Not the one you use?Change textbookSelect a different textbook
Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.75
Find the points on the curve y = 2x³ - 3x² - 12x + 20 where the tangent line is
a. perpendicular to the line y = 1 - (x/24).
_
b. parallel to the line y = √2 - 12x.
Verified step by step guidance1
To find the points where the tangent line is perpendicular to the line y = 1 - (x/24), first determine the slope of this line. The slope is -1/24.
For perpendicularity, the slope of the tangent line to the curve y = 2x³ - 3x² - 12x + 20 must be the negative reciprocal of -1/24, which is 24.
Find the derivative of the curve y = 2x³ - 3x² - 12x + 20 to get the slope of the tangent line: \( \frac{dy}{dx} = 6x^2 - 6x - 12 \). Set this equal to 24 and solve for x.
To find the points where the tangent line is parallel to the line y = √2 - 12x, determine the slope of this line, which is -12.
Set the derivative \( \frac{dy}{dx} = 6x^2 - 6x - 12 \) equal to -12 and solve for x to find the points where the tangent line is parallel.
Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Derivative
The derivative of a function measures the rate at which the function's value changes at a given point. It is the slope of the tangent line to the curve at that point. To find points where the tangent line is perpendicular or parallel to another line, we need to calculate the derivative of the curve and set it equal to the appropriate slope derived from the given lines.
Recommended video:
05:44
Derivatives
Slope of a Line
The slope of a line is a measure of its steepness, calculated as the change in y over the change in x (rise over run). For a line in the form y = mx + b, 'm' represents the slope. When determining where the tangent line to the curve is parallel or perpendicular to another line, we compare the slopes: parallel lines have equal slopes, while perpendicular lines have slopes that are negative reciprocals of each other.
Recommended video:
05:13Slopes of Tangent Lines
Critical Points
Critical points occur where the derivative of a function is zero or undefined, indicating potential locations for local maxima, minima, or points of inflection. In this context, finding critical points helps identify where the tangent line to the curve has specific slopes, which is essential for solving the problem of finding points where the tangent is parallel or perpendicular to the given lines.
Recommended video:
04:50Critical Points
Related Practice
Textbook Question
In Exercises 41–58, find dy/dt.
y = (t⁻³/⁴ sin(t))⁴/³
250
views
Textbook Question
a. Graph the function
ƒ(x) = { x², -1 ≤ x < 0
{ -x², 0 ≤ x ≤ 1.
b. Is ƒ continuous at x = 0?
c. Is ƒ differentiable at x = 0?
Give reasons for your answers.
155
views
Textbook Question
Find the derivatives of the functions in Exercises 19–40.
s = (4 / 3π)sin(3t) + (4 / 5π)cos(5t)
235
views
Textbook Question
In Exercises 41–58, find dy/dt.
y = tan²(sin³(t))
235
views
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 surface area S = 6x² of a cube when the edge lengths change from x₀ to x₀ + dx
204
views