Problem 253
Question
The radius of a right circular cone is increasing at 3 \(\mathrm{cm} / \mathrm{min}\) whereas the height of the cone is decreasing at 2 \(\mathrm{cm} / \mathrm{min}\). Find the rate of change of the volume of the cone when the radius is \(13 \mathrm{~cm}\) and the height is \(18 \mathrm{~cm}\).
Step-by-Step Solution
Verified Answer
The rate of change of the volume is approximately \(355.33\pi \text{ cm}^3/ ext{min}\).
1Step 1: Understand the Formula for Volume of a Cone
The volume \(V\) of a right circular cone is given by the formula \(V = \frac{1}{3} \pi r^2 h\), where \(r\) is the radius, and \(h\) is the height of the cone.
2Step 2: Calculate the Derivative with Respect to Time
To find the rate of change of the volume, differentiate \(V = \frac{1}{3} \pi r^2 h\) with respect to time \(t\). Using the product rule:\[\frac{dV}{dt} = \frac{1}{3} \pi \left(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt}\right)\]
3Step 3: Substitute Given Values
Substitute the given values of \(r = 13\), \(h = 18\), \(\frac{dr}{dt} = 3\), and \(\frac{dh}{dt} = -2\) into the differentiated expression:\[\frac{dV}{dt} = \frac{1}{3} \pi \left(2 \times 13 \times 18 \times 3 + 13^2 \times (-2)\right)\]
4Step 4: Simplify the Expression
Calculate each part of the expression:- \(2 \times 13 \times 18 \times 3 = 1404\)- \(13^2 \times (-2) = -338\)Now substitute these into the formula:\[\frac{dV}{dt} = \frac{1}{3} \pi (1404 - 338)\]
5Step 5: Compute the Result
Simplify to obtain:\(1404 - 338 = 1066\).Therefore, substitute into the equation:\[\frac{dV}{dt} = \frac{1}{3} \pi \times 1066\]\[\frac{dV}{dt} = \frac{1066}{3} \pi \approx 355.33\pi\] {(decimal approximation may vary)}.
Key Concepts
volume of a conedifferentiationproduct rule
volume of a cone
The volume of a cone is an essential geometric concept that helps us understand the space a cone occupies.
The volume is determined by both the base area of the cone and its height. The formula to find the volume of a right circular cone is \[ V = \frac{1}{3} \pi r^2 h \]where:
The volume is determined by both the base area of the cone and its height. The formula to find the volume of a right circular cone is \[ V = \frac{1}{3} \pi r^2 h \]where:
- \(V\) is the volume
- \(r\) is the radius of the base of the cone
- \(h\) is the height of the cone
- \(\pi\) (pi) is a constant, approximately 3.14159
differentiation
Differentiation is a key mathematical tool used to investigate the rates at which things change. In this exercise, differentiation helps us determine how the volume of the cone changes over time given the changes in radius and height.
When we have a function that describes something of interest—such as the volume of a cone—and we know that certain variables in this function are changing over time, we use differentiation to find the rate of change of the function's value. This means taking the derivative of the function with respect to time, denoted as \( \frac{d}{dt} \).
For the volume \( V = \frac{1}{3} \pi r^2 h \), differentiation helps us express how fast this volume is changing as the radius and height change. This is especially important in problems where dimensions of a figure like a cone change dynamically over time.
When we have a function that describes something of interest—such as the volume of a cone—and we know that certain variables in this function are changing over time, we use differentiation to find the rate of change of the function's value. This means taking the derivative of the function with respect to time, denoted as \( \frac{d}{dt} \).
For the volume \( V = \frac{1}{3} \pi r^2 h \), differentiation helps us express how fast this volume is changing as the radius and height change. This is especially important in problems where dimensions of a figure like a cone change dynamically over time.
product rule
The product rule in differentiation is employed when you need to differentiate a function that is the product of two or more sub-functions. It's an essential rule to understand since many real-world applications involve rates of change in such composite functions.
In the formula for the volume of a cone, \( V = \frac{1}{3} \pi r^2 h \), the terms \( r^2 \) and \( h \) are multiplied together. The product rule states that if you have a product \( u(t)v(t) \), the derivative of the product with respect to time \( t \) is:
\[ \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \]
Applying this to our cone, we let \( u = r^2 \) and \( v = h \), so:
In the formula for the volume of a cone, \( V = \frac{1}{3} \pi r^2 h \), the terms \( r^2 \) and \( h \) are multiplied together. The product rule states that if you have a product \( u(t)v(t) \), the derivative of the product with respect to time \( t \) is:
\[ \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \]
Applying this to our cone, we let \( u = r^2 \) and \( v = h \), so:
- \( \frac{d}{dt}[r^2 h] = \frac{d}{dt}[r^2] \cdot h + r^2 \cdot \frac{d}{dt}[h] \)
Other exercises in this chapter
Problem 250
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