Given the equation below, find when , , , and .
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- 0. Functions4h 53m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation2h 18m
- 4. Derivatives of Exponential & Logarithmic Functions1h 16m
- 5. Applications of Derivatives2h 19m
- 6. Graphical Applications of Derivatives6h 0m
- 7. Antiderivatives & Indefinite Integrals48m
- 8. Definite Integrals4h 36m
- 9. Graphical Applications of Integrals1h 43m
- 10. Integrals of Inverse, Exponential, & Logarithmic Functions21m
- 11. Techniques of Integration2h 7m
- 12. Trigonometric Functions6h 54m
- Angles29m
- Trigonometric Functions on Right Triangles1h 8m
- Solving Right Triangles23m
- Trigonometric Functions on the Unit Circle1h 19m
- Graphs of Sine & Cosine46m
- Graphs of Other Trigonometric Functions32m
- Trigonometric Identities52m
- Derivatives of Trig Functions42m
- Integrals of Basic Trig Functions28m
- Integrals of Other Trig Functions10m
- 13: Intro to Differential Equations2h 23m
- 14. Sequences & Series2h 8m
- 15. Power Series2h 19m
- 16. Probability & Calculus45m
5. Applications of Derivatives
Related Rates
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A 15-foot plank leans against a vertical pole. The top of the plank begins to slide down the pole at a steady speed of 2 inches per second. How fast is the bottom of the plank moving away from the pole when it is 8 feet away from the base of the pole (in inches per second)?
A
dx/dt=1.58sin
B
dx/dt=3.46sin
C
dx/dt=5.04sin
D
dx/dt=3.17sin

1
Step 1: Begin by identifying the relationship between the variables. The problem involves a right triangle formed by the plank, the pole, and the ground. Let x represent the distance of the bottom of the plank from the base of the pole (in feet), and y represent the height of the top of the plank on the pole (in feet). The length of the plank is constant at 15 feet, so we can use the Pythagorean theorem: x^2 + y^2 = 15^2.
Step 2: Differentiate the Pythagorean equation with respect to time t to relate the rates of change of x and y. Using implicit differentiation, we get: 2x(dx/dt) + 2y(dy/dt) = 0. Simplify this to: x(dx/dt) + y(dy/dt) = 0.
Step 3: Solve for dx/dt, the rate at which the bottom of the plank is moving away from the pole. Rearrange the equation to isolate dx/dt: dx/dt = -(y(dy/dt))/x.
Step 4: Substitute the known values into the equation. The problem states that dy/dt = -2 inches per second (negative because the top of the plank is sliding down). Convert all measurements to the same unit (inches). Since 1 foot = 12 inches, x = 8 feet = 96 inches, and the length of the plank is 15 feet = 180 inches. Use the Pythagorean theorem to find y: y = sqrt(180^2 - 96^2).
Step 5: Plug the values of y, dy/dt, and x into the equation for dx/dt: dx/dt = -(y(dy/dt))/x. Simplify the expression to find the rate at which the bottom of the plank is moving away from the pole in inches per second.
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