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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
All textbooksBriggs 3rd EditionCh. 4 - Applications of the DerivativeProblem 4.5.54b
Chapter 4, Problem 4.5.54b

{Use of Tech} Basketball shot A basketball is shot with an initial velocity of v ft/s at an angle of 45° to the floor. The center of the basketball is 8 ft above the floor at a horizontal distance of 18 feet from the center of the basketball hoop when it is released. The height h (in feet) of the center of the basketball after it has traveled a horizontal distance of x feet is modeled by the function h(x) = 32x² / v² + x + 8 (see figure). <IMAGE>




b. During the flight of the basketball, show that the distance s from the center of the basketball to the front of the hoop is s = √ (x - 17.25)² + ( -(4x² / 81) + x - 2)² (Hint: The diameter of the basketball hoop is 18 inches.) 

Verified step by step guidance
1
First, understand the problem: We need to show that the distance s from the center of the basketball to the front of the hoop is given by the formula s = √((x - 17.25)² + (-(4x² / 81) + x - 2)²).
Recognize that the formula for s involves calculating the horizontal and vertical distances between the basketball and the hoop. The horizontal distance is given by (x - 17.25), where x is the horizontal distance traveled by the basketball.
The vertical distance is given by the expression (-(4x² / 81) + x - 2). This expression represents the difference in height between the basketball and the hoop as the basketball travels horizontally.
To derive the formula for s, use the Pythagorean theorem. The distance s is the hypotenuse of a right triangle where one leg is the horizontal distance (x - 17.25) and the other leg is the vertical distance (-(4x² / 81) + x - 2).
Combine these distances using the Pythagorean theorem: s = √((horizontal distance)² + (vertical distance)²) = √((x - 17.25)² + (-(4x² / 81) + x - 2)²). This confirms the given formula for s.

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Key Concepts

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

Projectile Motion

Projectile motion describes the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. It involves two components: horizontal motion at a constant velocity and vertical motion under constant acceleration due to gravity. Understanding this helps in analyzing the basketball's trajectory and its height at any given horizontal distance.
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Distance Formula

The distance formula is used to determine the distance between two points in a plane. It is derived from the Pythagorean theorem and is given by s = √((x2 - x1)² + (y2 - y1)²). In this context, it helps calculate the distance from the basketball to the hoop by considering the horizontal and vertical displacements.
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Trigonometric Functions

Trigonometric functions relate the angles of a triangle to the lengths of its sides. In projectile motion, they are used to resolve the initial velocity into horizontal and vertical components. For a launch angle of 45°, these components are equal, simplifying calculations of the basketball's path and its position at any point in time.
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Suppose S = x + 2y is an objective function subject to the constraint xy = 50, for x > 0 and y > 0.

b. Find the absolute minimum value of S subject to the given constraint.

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b. A rancher plans to make four identical and adjacent rectangular pens against a barn, each with an area of 100 m² (see figure). What are the dimensions of each pen that minimize the amount of fence that must be used? <IMAGE>

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Values of related functions Suppose f is differentiable on (-∞,∞) and assume it has a local extreme value at the point x = 2, where f(2) = 0. Let g(x) = xf(x) + 1 and let h(x) = xf(x) + x +1, for all values of x.


b. Does either g or h have a local extreme value at x = 2? Explain.

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{Use of Tech} A damped oscillator The displacement of an object as it bounces vertically up and down on a spring is given by y(t) = 2.5e⁻ᵗ cos 2t, where the initial displacement is y(0) = 2.5 and y = 0 corresponds to the rest position (see figure). <IMAGE>

b. Find the time and the displacement when the object reaches its lowest point.

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Cylinder in a cone A right circular cylinder is placed inside a cone of radius R and height H so that the base of the cylinder lies on the base of the cone.


b. Find the dimensions of the cylinder with maximum lateral surface area (area of the curved surface).

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Two poles of heights m and n are separated by a horizontal distance d. A rope is stretched from the top of one pole to the ground and then to the top of the other pole. Show that the configuration that requires the least amount of rope occurs when Θ₁ = Θ₂ (see figure). <IMAGE>

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