Ch. 3 - Derivatives

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 3 - Derivatives

Problem 3.6.23c

Chapter 3, Problem 3.6.23c

Throwing a stone Suppose a stone is thrown vertically upward from the edge of a cliff on Earth with an initial velocity of 32 ft/s from a height of 48 ft above the ground. The height (in feet) of the stone above the ground t seconds after it is thrown is s(t) = -16t²+32t+48.
c. What is the height of the stone at the highest point?

Verified step by step guidance

To find the height of the stone at its highest point, we need to determine when the stone reaches its maximum height. This occurs at the vertex of the parabola represented by the quadratic function s(t) = -16t² + 32t + 48.

The vertex of a parabola given by the equation ax² + bx + c can be found using the formula t = -b/(2a). In this case, a = -16 and b = 32.

Substitute the values of a and b into the vertex formula: t = -32/(2 * -16). This will give you the time t at which the stone reaches its maximum height.

Once you have the value of t, substitute it back into the original height function s(t) = -16t² + 32t + 48 to find the height of the stone at this time.

Evaluate s(t) using the calculated value of t to determine the maximum height of the stone above the ground.

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Key Concepts

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

Quadratic Functions

The height of the stone is modeled by a quadratic function, which is a polynomial of degree two. Quadratic functions have a parabolic shape and can be expressed in the form s(t) = at² + bt + c, where a, b, and c are constants. The vertex of the parabola represents the maximum or minimum point, which is crucial for determining the highest point of the stone's trajectory.

Vertex of a Parabola

The vertex of a parabola given by the function s(t) = -16t² + 32t + 48 can be found using the formula t = -b/(2a). In this case, 'a' is -16 and 'b' is 32. The vertex provides the time at which the stone reaches its maximum height, and substituting this time back into the height function gives the maximum height.

Maximizing Functions

To find the maximum height of the stone, we need to evaluate the function at the vertex. This involves calculating the height at the time derived from the vertex formula. Understanding how to maximize a function is essential in calculus, as it applies to various real-world scenarios, including projectile motion like that of the thrown stone.

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Textbook Question

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.

c. What is the height of the stone at the highest point?

Textbook Question

Deriving trigonometric identities

c. Differentiate both sides of the identity sin 2t = 2 sin t cost to prove that cos 2t = cos²t−sin²t.

Textbook Question

Derivatives using tables Let $h(x)=f(g(x))$ and $p(x)=g(f(x))$ . Use the table to compute the following derivatives.

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c. $p prime of 4$

Textbook Question

City urbanization City planners model the size of their city using the function A(t) = - 1/50t² + 2t +20, for 0 ≤ t ≤ 50, where A is measured in square miles and t is the number of years after 2010.

c. Suppose the population density of the city remains constant from year to year at 1000 people mi². Determine the growth rate of the population in 2030.

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Textbook Question

Vibrations of a spring Suppose an object of mass m is attached to the end of a spring hanging from the ceiling. The mass is at its equilibrium position $y equals 0$ when the mass hangs at rest. Suppose you push the mass to a position $y sub 0$ units above its equilibrium position and release it. As the mass oscillates up and down (neglecting any friction in the system), the position y of the mass after t seconds is $y = y_{0} \cos (t \sqrt{\frac{k}{m}})$ , where $k is greater than 0$ is a constant measuring the stiffness of the spring (the larger the value of $k$ , the stiffer the spring) and $y$ is positive in the upward direction.

Use equation (4) to answer the following questions.

c. How would the velocity be affected if the experiment were repeated with a spring having four times the stiffness ( $k$ is increased by a factor of $4$ )?

Textbook Question

Witch of Agnesi Let y(x²+4)=8 (see figure). <IMAGE>

c. Solve the equation y(x²+4)=8 for y to find an explicit expression for y and then calculate dy/dx.