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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 3 - DerivativesProblem 25f
Chapter 3, Problem 25f

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 64 ft/s from a height of 32 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t2 + 64t + 32.
On what intervals is the speed increasing?

Verified step by step guidance
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First, understand that the speed of the stone is the absolute value of its velocity. To find when the speed is increasing, we need to determine when the velocity is increasing in magnitude.
The velocity function v(t) is the derivative of the height function s(t). Calculate the derivative: v(t) = d/dt [-16t^2 + 64t + 32].
Compute the derivative: v(t) = -32t + 64. This represents the velocity of the stone at any time t.
To find when the speed is increasing, we need to determine when the magnitude of v(t) is increasing. This involves finding when the acceleration, which is the derivative of the velocity, is positive.
Calculate the acceleration: a(t) = d/dt [v(t)] = d/dt [-32t + 64] = -32. Since the acceleration is constant and negative, the speed is not increasing at any interval.

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Key Concepts

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

Velocity and Speed

Velocity is a vector quantity that refers to the rate of change of position with respect to time, including direction. Speed, on the other hand, is the magnitude of velocity and does not consider direction. In this context, understanding how velocity changes over time is crucial for determining when the speed of the stone is increasing.
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Derivatives Applied To Velocity

Acceleration

Acceleration is the rate of change of velocity with respect to time. In the given function for height, the acceleration is constant and negative due to gravity, which affects the stone's motion. To find intervals where speed is increasing, one must analyze when the acceleration is positive, indicating that the velocity is increasing in the positive direction.
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Critical Points and Intervals

Critical points occur where the derivative of a function is zero or undefined, indicating potential maxima, minima, or points of inflection. To determine intervals where speed is increasing, one must find the derivative of the velocity function and analyze the sign of this derivative across different intervals. This helps identify where the speed transitions from decreasing to increasing.
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Related Practice
Textbook Question

Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 19.6 m/s from a height of 24.5 m above the ground. The height (in meters) of the stone above the ground t seconds after it is thrown is s(t) = -4.9t2 + 19.6t + 24.5.

On what intervals is the speed increasing?

Textbook Question

Suppose a baseball is thrown vertically upward from the ground with an initial velocity of v0 ft/s. The approximate height of the ball (in feet) above the ground after t seconds is given by s(t) = -16t² + v0t.

With what velocity does the ball strike the ground?

Textbook Question

Once Kate’s kite reaches a height of 50 ft (above her hands), it rises no higher but drifts due east in a wind blowing 5 ft/s. How fast is the string running through Kate’s hands at the moment when she has released 120 ft of string?

Textbook Question

Evaluate the derivative of the following functions.

f(t) = ln (sin-1 t2)

1
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Textbook Question

Use definition (2) (p. 135) to find the slope of the line tangent to the graph of f at P.

f(x) = x3; P (1,1)

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Textbook Question

Evaluate the derivative of the following functions.

f(x) = x2 + 2x3 cot-1 x - ln (1 + x2)