Textbook Question
6. You are planning to close off a corner of the first quadrant with a line segment 20 units long running from (a, 0) to (0,b). Show that the area of the triangle enclosed by the segment is largest when a = b.
Ch. 4 - Applications of Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 4, Problem 4.1.61
Theory and Examples
Maximum height of a vertically moving body The height of a body moving vertically is given by s = −12gt² + υ₀t + s₀, g > 0, with s in meters and t in seconds. Find the body’s maximum height.
Verified step by step guidance1
Identify the given equation for the height of the body: \( s(t) = -12gt^2 + v_0t + s_0 \), where \( g \) is the acceleration due to gravity, \( v_0 \) is the initial velocity, and \( s_0 \) is the initial height.
To find the maximum height, we need to determine the time \( t \) at which the velocity is zero. The velocity \( v(t) \) is the derivative of the height function \( s(t) \) with respect to time \( t \).
Differentiate the height function: \( v(t) = \frac{d}{dt}(-12gt^2 + v_0t + s_0) = -24gt + v_0 \).
Set the velocity \( v(t) \) to zero to find the critical point: \( -24gt + v_0 = 0 \). Solve for \( t \) to find \( t = \frac{v_0}{24g} \).
Substitute \( t = \frac{v_0}{24g} \) back into the original height equation \( s(t) \) to find the maximum height: \( s_{max} = -12g\left(\frac{v_0}{24g}\right)^2 + v_0\left(\frac{v_0}{24g}\right) + s_0 \). Simplify this expression to find the maximum height.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Quadratic Functions
A quadratic function is a polynomial of degree two, typically in the form of ax² + bx + c. The graph of a quadratic function is a parabola, which can open upwards or downwards. In the context of the problem, the height function s = −12gt² + υ₀t + s₀ is a quadratic function of time t, where the coefficient of t² determines the direction of the parabola.
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Introduction to Polynomial Functions
Vertex of a Parabola
The vertex of a parabola is the point where it reaches its maximum or minimum value. For a quadratic function in the form ax² + bx + c, the vertex can be found using the formula t = -b/(2a). In this problem, the vertex represents the time at which the body reaches its maximum height, as the parabola opens downwards due to the negative coefficient of t².
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7:42Properties of Parabolas
Derivative and Critical Points
The derivative of a function gives the rate of change and can be used to find critical points where the function's slope is zero. These points are potential maxima or minima. For the height function, taking the derivative with respect to time and setting it to zero helps find the time at which the maximum height occurs, confirming the vertex calculation.
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04:50Critical Points
Related Practice
Textbook Question
Initial Value Problems
Solve the initial value problems in Exercises 71–90.
dy/dx = 2x − 7, y(2) = 0
Textbook Question
Finding Position from Velocity or Acceleration
Exercises 45–48 give the acceleration a=d²s/dt², initial velocity, and initial position of an object moving on a coordinate line. Find the object’s position at time t.
a = 9.8, v(0) = −3, s(0) = 0
Textbook Question
Finding Indefinite Integrals
In Exercises 17–56, find the most general antiderivative or indefinite integral. You may need to try a solution and then adjust your guess. Check your answers by differentiation.
∫(cscθ cotθ) / 2 dθ
Textbook Question
[Technology Exercises] When solving Exercises 14–30, you may need to use appropriate technology (such as a calculator or a computer).
26. Factoring a quartic Find the approximate values of r_1 through r_4 in the factorization
8x^4-14x^3-9x^2+11x-1=8(x-r_1)(x-r_2)(x-r_3)(x-r_4)
Textbook Question
Theory and Examples
[Technology Exercise] Graph the functions in Exercises 63–66. Then find the extreme values of the function on the interval and say where they occur.
f(x) = |x − 2| + |x + 3|, −5 ≤ x ≤ 5