Problem 38
Question
Approximate the change in the volume of a right circular cone of fixed height \(h=4 \mathrm{m}\) when its radius increases from \(r=3 \mathrm{m}\) to \(r=3.05 \mathrm{m}\left(V(r)=\pi r^{2} h / 3\right)\).
Step-by-Step Solution
Verified Answer
Answer: The approximate change in volume is around \(0.4033\bar{3}\pi\) cubic meters.
1Step 1: Understand the problem
First, we need to understand the dimensions of the cone. It has a fixed height of 4 meters and its radius is to be increased from 3 meters to 3.05 meters. We will need to find the volume before and after this increase in radius, and then find the difference between these volumes.
2Step 2: Recall the formula for the volume of a cone
The formula to find the volume of a cone is given by:
\(V(r) = \frac{1}{3} \pi r^2 h\)
In our problem, \(h = 4\) meters is fixed, and we will first find the volume when the radius is \(3\) meters, and then find the volume when the radius is \(3.05\) meters.
3Step 3: Find the initial volume
Substitute the given values into the formula to find the initial volume:
\(V(3) = \frac{1}{3} \pi (3)^2 (4) = \frac{1}{3} \pi (9)(4)\)
So, \(V(3) = 12\pi\) cubic meters
4Step 4: Find the final volume
Next, find the volume after the increase in radius to \(3.05\) meters:
\(V(3.05) = \frac{1}{3} \pi (3.05)^2 (4) = \frac{1}{3} \pi (9.3025)(4)\)
So, \(V(3.05) = 12.4033\bar{3} \pi\) cubic meters
5Step 5: Calculate the change in volume
Now, find the difference between the final and initial volumes:
\(\Delta V = V(3.05) - V(3) = 12.4033\bar{3}\pi - 12\pi\)
\(\Delta V \approx 0.4033\bar{3}\pi\) cubic meters
The approximate change in the volume of the cone when its radius increases from \(3\) meters to \(3.05\) meters is around \(0.4033\bar{3}\pi\) cubic meters.
Key Concepts
Approximate ChangeVolume of a ConeRadius Increase
Approximate Change
When we talk about approximate change, we are looking at how a small change in one variable affects the output of a function. In calculus, this is often addressed using derivatives. In this exercise, we want to see how a small change in the radius of a cone affects its volume. The formula we use relates the radius to the volume. We calculate the volume at two slightly different radii and find the difference.
For our cone, with a very small increase in radius from 3 to 3.05 meters, we want to approximate the change in volume:
For our cone, with a very small increase in radius from 3 to 3.05 meters, we want to approximate the change in volume:
- The smaller the change in the radius, the closer the approximation will be to the real change in volume.
- This method offers an efficient way to predict changes with minimal computation, especially when dealing with nonlinear relationships.
Volume of a Cone
Understanding how to compute the volume of a cone is crucial. The formula used is:\[ V(r) = \frac{1}{3} \pi r^2 h \]
This formula calculates the space inside a right circular cone. Here’s what each symbol represents:
With this formula, you can determine how much space is inside the cone as its radius changes. Multiplying \( \pi r^2 \) by the height and dividing by 3 gives you the volume. For our problem:
This formula calculates the space inside a right circular cone. Here’s what each symbol represents:
- \( \pi \): A constant approximately equal to 3.14159, representing the ratio of a circle's circumference to its diameter.
- \( r \): The radius of the circular base of the cone.
- \( h \): The height of the cone, which is constant in our exercise.
With this formula, you can determine how much space is inside the cone as its radius changes. Multiplying \( \pi r^2 \) by the height and dividing by 3 gives you the volume. For our problem:
- Initial calculation with \( r = 3 \) meters gives \( 12 \pi \) cubic meters.
- After increasing the radius to 3.05 meters, the volume becomes \( 12.4033\bar{3} \pi \) cubic meters.
Radius Increase
Increasing the radius of a cone affects its volume rather significantly. This is because the volume of a cone is proportional to the square of the radius. Here's how it breaks down:
This amplification effect makes understanding how small changes in dimension impact overall measurements in volume important. By noting the approximate volume change \( 0.4033\bar{3} \pi \) cubic meters, we see the impact of such a modest increase in radius.
- When the radius increases, even slightly, the squared term \( r^2 \) increases the resulting volume quite a bit because squaring amplifies the change.
- In our specific example, increasing the radius from 3 to 3.05 meters might seem small, but due to the squared radius term in the formula, the change in volume is noticeable.
This amplification effect makes understanding how small changes in dimension impact overall measurements in volume important. By noting the approximate volume change \( 0.4033\bar{3} \pi \) cubic meters, we see the impact of such a modest increase in radius.
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