Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

4. Applications of Derivatives

Related Rates

Calculus

4. Applications of Derivatives

Related Rates

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2:36 minutes

Problem 3.8.6

Textbook Question

If x = y³ – y and dy/dt = 5, then what is dx/dt when y = 2?

Verified step by step guidance

First, recognize that you need to find the rate of change of x with respect to time, dx/dt, given the equation x = y³ - y and the rate of change of y with respect to time, dy/dt = 5.

To find dx/dt, use implicit differentiation with respect to time t on both sides of the equation x = y³ - y. This involves applying the chain rule.

Differentiate the left side of the equation with respect to t: d(x)/d(t) = dx/dt.

Differentiate the right side of the equation with respect to t: d(y³ - y)/d(t) = 3y²(dy/dt) - (dy/dt).

Substitute dy/dt = 5 and y = 2 into the differentiated equation to solve for dx/dt: dx/dt = 3(2)²(5) - (5).

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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 find the derivative of a function when it is not explicitly solved for one variable in terms of another. In this problem, x is given in terms of y, and we need to differentiate both sides with respect to time t, applying the chain rule to account for dy/dt.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Here, it helps relate dx/dt to dy/dt by differentiating x = y³ - y with respect to t.

Substitution

Substitution involves replacing variables with known values to simplify expressions or solve equations. In this problem, after finding the expression for dx/dt in terms of y and dy/dt, we substitute y = 2 and dy/dt = 5 to calculate the specific value of dx/dt at that point.

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Textbook Question

Economics

Marginal revenue

Suppose that the revenue from selling x washing machines is

r(x) = 20000(1 − 1/x) dollars.

b. Use the function r'(x) to estimate the increase in revenue that will result from increasing production from 100 machines a week to 101 machines a week.