Textbook Question
Find an equation of the straight line having slope 1/4 that is tangent to the curve y = √x.
Ch. 3 - 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 3, Problem 3.8.27
A growing sand pile Sand falls from a conveyor belt at the rate of 10 m³/min onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the (a) height and (b) radius changing when the pile is 4 m high? Answer in centimeters per minute.
Verified step by step guidance1
First, express the relationship between the height (h) and the radius (r) of the cone. Since the height is three-eighths of the base diameter, we have h = (3/8) * 2r, which simplifies to h = (3/4)r.
Next, use the formula for the volume of a cone, V = (1/3)πr²h. Substitute h = (3/4)r into this formula to express the volume in terms of r: V = (1/3)πr²((3/4)r) = (1/4)πr³.
Differentiate the volume with respect to time (t) to find the rate of change of the volume: dV/dt = (1/4)π * 3r² * dr/dt. Given that dV/dt = 10 m³/min, substitute this value into the equation.
Solve for dr/dt, the rate of change of the radius, when the height is 4 m. Since h = (3/4)r, when h = 4 m, r = 4 * (4/3) = 16/3 m. Substitute r = 16/3 m into the differentiated volume equation to find dr/dt.
Finally, use the relationship h = (3/4)r to find dh/dt in terms of dr/dt. Since h = (3/4)r, differentiate both sides with respect to time to get dh/dt = (3/4) * dr/dt. Substitute the value of dr/dt to find dh/dt in centimeters per minute.
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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 the height and radius of the sand pile change over time as sand is added. This requires setting up equations that relate these quantities and differentiating with respect to time.
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Intro To Related Rates
Volume of a Cone
The volume of a cone is given by the formula V = (1/3)πr²h, where r is the radius and h is the height. Understanding this formula is crucial because the rate at which sand is added affects the volume, and we need to relate this to changes in the height and radius of the cone.
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Geometric Relationships
The problem states that the height of the pile is always three-eighths of the base diameter, which implies a specific relationship between the height and radius: h = (3/4)r. This relationship is essential for expressing one variable in terms of the other, simplifying the differentiation process needed to find the rates of change.
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Related Practice
Textbook Question
Find dy/dt when x = 1 if y = x² + 7x − 5 and dx/dt = ¹/₃.
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Theory and Examples
Intersecting normal line The line that is normal to the curve x² + 2xy – 3y² = 0 at (1,1) intersects the curve at what other point?
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In Exercises 41–58, find dy/dt.
y = √(3t + (√2 + √(1 − t)))
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Find the derivatives of the functions in Exercises 17–28.
y = ((x + 1)(x + 2)) / ((x − 1)(x − 2))
Textbook Question
If r + s² + v³ = 12, dr/dt = 4, and ds/dt = –3, find dv/dt when r = 3 and s = 1.