A balloon and a bicycle A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec. Just when the balloon is 65 ft above the ground, a bicycle moving at a constant rate of 17 ft/sec passes under it. How fast is the distance s(t) between the bicycle and the balloon increasing 3 sec later?
Table of contents
- 0. Functions7h 55m
- Introduction to Functions18m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms36m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 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
- 10. Physics Applications of Integrals 3h 16m
- 11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 31m
- 12. Techniques of Integration7h 41m
- 13. Intro to Differential Equations2h 55m
- 14. Sequences & Series5h 36m
- 15. Power Series2h 19m
- 16. Parametric Equations & Polar Coordinates7h 58m
4. Applications of Derivatives
Related Rates
Multiple Choice
Given the equation , what is the value of at the point ?
A
B
C
D
0 Comments
Verified step by step guidance1
Step 1: Recognize that the problem involves implicit differentiation since the equation contains both x and y variables. The goal is to find dy/dx at the given point (2, 2).
Step 2: Differentiate both sides of the equation e^x - y = x y^3 + e^2 - 18 with respect to x. Remember to apply the chain rule when differentiating terms involving y, as y is a function of x.
Step 3: For the left-hand side, differentiate e^x to get e^x and differentiate -y to get -dy/dx. For the right-hand side, use the product rule on x y^3, which gives (d/dx)(x y^3) = y^3 + x * 3y^2 * dy/dx. The derivative of constants like e^2 and -18 is 0.
Step 4: Combine all differentiated terms into a single equation: e^x - dy/dx = y^3 + x * 3y^2 * dy/dx. Rearrange the equation to isolate dy/dx.
Step 5: Substitute the point (2, 2) into the equation to evaluate dy/dx. Plug in x = 2 and y = 2, and solve for dy/dx algebraically.
Related Videos
Related Practice
Textbook Question
200
views

