Two measurements of a cylinder are varying in such a way that the volume is kept constant. If the rates of change of the radius (r ) and height (h) are equal in magnitude but opposite in sign, then

Options

A. r=2h
B. h=2r
C. h=r
D. h=2r

Explanation

To solve the problem, we need to analyze the relationship between the radius \( r \) and height \( h \) of a cylinder when its volume remains constant while both \( r \) and \( h \) are changing. Let's go through the solution step by step.

### Step 1: Understand the Volume of a Cylinder
The volume \( V \) of a cylinder is given by the formula:
\[ V = \pi r^2 h \]
where \( r \) is the radius and \( h \) is the height.

### Step 2: Differentiate the Volume with Respect to Time
Since the volume is constant, the rate of change of volume with respect to time \( \frac{dV}{dt} \) is zero:
\[ \frac{dV}{dt} = 0 \]
Now, we differentiate the volume formula using the product rule:
\[ \frac{dV}{dt} = \pi \left( \frac{d}{dt}(r^2) \cdot h + r^2 \cdot \frac{dh}{dt} \right) \]

### Step 3: Differentiate \( r^2 \)
Using the chain rule, we differentiate \( r^2 \):
\[ \frac{d}{dt}(r^2) = 2r \frac{dr}{dt} \]
Thus, we can substitute this back into our differentiation of volume:
\[ \frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} \cdot h + r^2 \frac{dh}{dt} \right) \]

### Step 4: Set the Rate of Change of Volume to Zero
Since \( \frac{dV}{dt} = 0 \):
\[ 0 = \pi \left( 2r h \frac{dr}{dt} + r^2 \frac{dh}{dt} \right) \]
We can simplify this equation by dividing both sides by \( \pi \) (since \( \pi \neq 0 \)):
\[ 0 = 2r h \frac{dr}{dt} + r^2 \frac{dh}{dt} \]

### Step 5: Substitute the Relationship Between \( \frac{dr}{dt} \) and \( \frac{dh}{dt} \)
According to the problem, the rates of change of the radius and height are equal in magnitude but opposite in sign:
\[ \frac{dr}{dt} = -\frac{dh}{dt} \]
Substituting this into our equation gives:
\[ 0 = 2r h \frac{dr}{dt} - r^2 \frac{dr}{dt} \]

### Step 6: Factor Out \( \frac{dr}{dt} \)
Factoring out \( \frac{dr}{dt} \):
\[ \frac{dr}{dt} (2rh - r^2) = 0 \]
Since \( \frac{dr}{dt} \) cannot be zero (otherwise, the radius wouldn't be changing), we set the other factor to zero:
\[ 2rh - r^2 = 0 \]

### Step 7: Solve for the Relationship Between \( r \) and \( h \)
Rearranging gives:
\[ r^2 = 2rh \]
Dividing both sides by \( r \) (assuming \( r \neq 0 \)):
\[ r = 2h \]

### Conclusion
Thus, the relationship that follows is:
\[ r = 2h \]