Question
Glass A - Cylinder, Glass B - Cone: The water glasses are filled to the same height and have the same radius. Fill in the blanks: It would take _____times filling up glass B and pouring into glass A to fill up glass A; If the height of the glasses doubled, then the volume would increase times____. If the radius of the glasses doubled, then the volumes would increase times ____; If the radius and the height both doubled, then the volumes would increase times ____. Use your formula to substitute V = 1/3 pi r2 and V = pi r2 h ( one; two; three; four; five; six; seven; eight; nine)
Answers
GPT-4o mini
Let's analyze the volumes of the two types of glasses using the provided formulas.
1. **Volume of Glass A (Cylinder)**:
\[
V_A = \pi r^2 h
\]
2. **Volume of Glass B (Cone)**:
\[
V_B = \frac{1}{3} \pi r^2 h
\]
### Part 1: How many times filling Glass B to fill Glass A
To find out how many times we need to fill glass B to fill glass A, divide the volume of glass A by the volume of glass B:
\[
\text{Number of fills} = \frac{V_A}{V_B} = \frac{\pi r^2 h}{\frac{1}{3} \pi r^2 h}
\]
\[
= \frac{\pi r^2 h \cdot 3}{\pi r^2 h} = 3
\]
It would take **3** times filling up glass B and pouring into glass A to fill up glass A.
### Part 2: If the height of the glasses is doubled
If the height is doubled, the new height \( h' = 2h \). The volumes of the glasses become:
- For glass A (cylinder):
\[
V_A' = \pi r^2 (2h) = 2\pi r^2 h
\]
- For glass B (cone):
\[
V_B' = \frac{1}{3} \pi r^2 (2h) = \frac{2}{3} \pi r^2 h
\]
The volume of both glasses increases proportionally based on the height.
Thus, the volume increases:
\[
\text{Cylindrical Volume Factor} = \frac{V_A'}{V_A} = \frac{2\pi r^2 h}{\pi r^2 h} = 2
\]
\[
\text{Conical Volume Factor} = \frac{V_B'}{V_B} = \frac{\frac{2}{3}\pi r^2 h}{\frac{1}{3}\pi r^2 h} = 2
\]
The volume would increase times **2**.
### Part 3: If the radius of the glasses is doubled
If the radius is doubled, the new radius \( r' = 2r \). The volumes of the glasses become:
- For glass A (cylinder):
\[
V_A' = \pi (2r)^2 h = 4\pi r^2 h
\]
- For glass B (cone):
\[
V_B' = \frac{1}{3} \pi (2r)^2 h = \frac{4}{3} \pi r^2 h
\]
So, the volume increases by a factor of:
\[
\text{Cylindrical Volume Factor} = \frac{V_A'}{V_A} = \frac{4\pi r^2 h}{\pi r^2 h} = 4
\]
\[
\text{Conical Volume Factor} = \frac{V_B'}{V_B} = \frac{\frac{4}{3}\pi r^2 h}{\frac{1}{3}\pi r^2 h} = 4
\]
The volume would increase times **4**.
### Part 4: If both the radius and height are doubled
If both the radius and height are doubled, the new radius \( r' = 2r \) and new height \( h' = 2h \). The volumes become:
- For glass A (cylinder):
\[
V_A' = \pi (2r)^2 (2h) = 8\pi r^2 h
\]
- For glass B (cone):
\[
V_B' = \frac{1}{3} \pi (2r)^2 (2h) = \frac{8}{3} \pi r^2 h
\]
So, the volume increases by a factor of:
\[
\text{Cylindrical Volume Factor} = \frac{V_A'}{V_A} = \frac{8\pi r^2 h}{\pi r^2 h} = 8
\]
\[
\text{Conical Volume Factor} = \frac{V_B'}{V_B} = \frac{\frac{8}{3}\pi r^2 h}{\frac{1}{3}\pi r^2 h} = 8
\]
The volume would increase times **8**.
### Summary:
1. It would take **3** times filling up glass B and pouring into glass A to fill up glass A.
2. If the height of the glasses doubled, then the volume would increase times **2**.
3. If the radius of the glasses doubled, then the volumes would increase times **4**.
4. If the radius and the height both doubled, then the volumes would increase times **8**.
1. **Volume of Glass A (Cylinder)**:
\[
V_A = \pi r^2 h
\]
2. **Volume of Glass B (Cone)**:
\[
V_B = \frac{1}{3} \pi r^2 h
\]
### Part 1: How many times filling Glass B to fill Glass A
To find out how many times we need to fill glass B to fill glass A, divide the volume of glass A by the volume of glass B:
\[
\text{Number of fills} = \frac{V_A}{V_B} = \frac{\pi r^2 h}{\frac{1}{3} \pi r^2 h}
\]
\[
= \frac{\pi r^2 h \cdot 3}{\pi r^2 h} = 3
\]
It would take **3** times filling up glass B and pouring into glass A to fill up glass A.
### Part 2: If the height of the glasses is doubled
If the height is doubled, the new height \( h' = 2h \). The volumes of the glasses become:
- For glass A (cylinder):
\[
V_A' = \pi r^2 (2h) = 2\pi r^2 h
\]
- For glass B (cone):
\[
V_B' = \frac{1}{3} \pi r^2 (2h) = \frac{2}{3} \pi r^2 h
\]
The volume of both glasses increases proportionally based on the height.
Thus, the volume increases:
\[
\text{Cylindrical Volume Factor} = \frac{V_A'}{V_A} = \frac{2\pi r^2 h}{\pi r^2 h} = 2
\]
\[
\text{Conical Volume Factor} = \frac{V_B'}{V_B} = \frac{\frac{2}{3}\pi r^2 h}{\frac{1}{3}\pi r^2 h} = 2
\]
The volume would increase times **2**.
### Part 3: If the radius of the glasses is doubled
If the radius is doubled, the new radius \( r' = 2r \). The volumes of the glasses become:
- For glass A (cylinder):
\[
V_A' = \pi (2r)^2 h = 4\pi r^2 h
\]
- For glass B (cone):
\[
V_B' = \frac{1}{3} \pi (2r)^2 h = \frac{4}{3} \pi r^2 h
\]
So, the volume increases by a factor of:
\[
\text{Cylindrical Volume Factor} = \frac{V_A'}{V_A} = \frac{4\pi r^2 h}{\pi r^2 h} = 4
\]
\[
\text{Conical Volume Factor} = \frac{V_B'}{V_B} = \frac{\frac{4}{3}\pi r^2 h}{\frac{1}{3}\pi r^2 h} = 4
\]
The volume would increase times **4**.
### Part 4: If both the radius and height are doubled
If both the radius and height are doubled, the new radius \( r' = 2r \) and new height \( h' = 2h \). The volumes become:
- For glass A (cylinder):
\[
V_A' = \pi (2r)^2 (2h) = 8\pi r^2 h
\]
- For glass B (cone):
\[
V_B' = \frac{1}{3} \pi (2r)^2 (2h) = \frac{8}{3} \pi r^2 h
\]
So, the volume increases by a factor of:
\[
\text{Cylindrical Volume Factor} = \frac{V_A'}{V_A} = \frac{8\pi r^2 h}{\pi r^2 h} = 8
\]
\[
\text{Conical Volume Factor} = \frac{V_B'}{V_B} = \frac{\frac{8}{3}\pi r^2 h}{\frac{1}{3}\pi r^2 h} = 8
\]
The volume would increase times **8**.
### Summary:
1. It would take **3** times filling up glass B and pouring into glass A to fill up glass A.
2. If the height of the glasses doubled, then the volume would increase times **2**.
3. If the radius of the glasses doubled, then the volumes would increase times **4**.
4. If the radius and the height both doubled, then the volumes would increase times **8**.