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Ch. 3 - Derivatives
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 3 - DerivativesProblem 3.8.33
Chapter 3, Problem 3.8.33

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?
A hot air balloon rises vertically while a bicycle moves horizontally below it, illustrating related rates in calculus.

Verified step by step guidance
1
Identify the variables: Let y(t) be the height of the balloon above the ground, x(t) be the horizontal distance of the bicycle from the point directly below the balloon, and s(t) be the distance between the balloon and the bicycle.
Write the given rates: The balloon rises at a rate of 1 ft/sec, so dy/dt = 1 ft/sec. The bicycle moves at a rate of 17 ft/sec, so dx/dt = 17 ft/sec.
Use the Pythagorean theorem to relate the distances: s(t)^2 = x(t)^2 + y(t)^2. Differentiate both sides with respect to time t to find the rate of change of s(t).
Differentiate: 2s(t) * ds/dt = 2x(t) * dx/dt + 2y(t) * dy/dt. Simplify to find ds/dt, the rate at which the distance between the balloon and the bicycle is increasing.
Substitute the known values: At t = 3 seconds, y(t) = 65 + 3*1 = 68 ft, x(t) = 3*17 = 51 ft. Use these values to solve for ds/dt.

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Key Concepts

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

Related Rates

Related rates involve finding the rate at which one quantity changes with respect to another. In this problem, we need to determine how fast the distance between the balloon and the bicycle changes over time. This requires understanding how the rates of change of the vertical and horizontal positions relate to the rate of change of the distance between them.
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Pythagorean Theorem

The Pythagorean Theorem is essential for relating the vertical and horizontal distances to the hypotenuse, which is the distance s(t) between the balloon and the bicycle. Given the vertical distance y(t) and horizontal distance x(t), the theorem states that s(t)^2 = y(t)^2 + x(t)^2, allowing us to express s(t) in terms of y(t) and x(t).
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Differentiation

Differentiation is used to find the rate of change of a function. In this context, we differentiate the equation s(t)^2 = y(t)^2 + x(t)^2 with respect to time to find ds/dt, the rate at which the distance between the balloon and the bicycle changes. This involves applying the chain rule to differentiate each term with respect to time.
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Related Practice
Textbook Question

Second Derivatives


In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.


x²/³ + y²/³ = 1

Textbook Question

Find the first and second derivatives of the functions in Exercises 33–38.


w = ((1 + 3z) / 3z) (3 − z)

Textbook Question

Find the derivatives of the functions in Exercises 1–42.

_____

𝔂 = / x² + x

√ x²

1
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Textbook Question

One-Sided Derivatives


Compute the right-hand and left-hand derivatives as limits to show that the functions in Exercises 37–40 are not differentiable at the point P.

Textbook Question

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In Exercises 19–26, use implicit differentiation to find dy/dx and then d²y/dx². Write the solutions in terms of x and y only.


y² – 2x = 1 – 2y

Textbook Question

In Exercises 51 and 52, find dp/dq.


p³ + 4pq - 3q² = 2