Suppose that water is emptied from a spherical tank of radius 10 cm. If the depth of the water in the tank is 4 cm and is decreasing at the rate of 2 cm/sec, then the radius of the top surface of water is decreasing at the rate of

Options

A. 1
B. `2//3`
C. `-3//2`
D. 2

Explanation

To solve the problem, we need to find the rate at which the radius of the top surface of the water (denoted as \( x \)) is decreasing as the water is emptied from a spherical tank. Here are the steps to derive the solution:

### Step 1: Understand the Geometry
We have a spherical tank with a radius of \( R = 10 \) cm. The depth of the water in the tank is \( h = 4 \) cm, and it is decreasing at a rate of \( \frac{dh}{dt} = -2 \) cm/sec (the negative sign indicates that the height is decreasing).

### Step 2: Relate the Variables
Using the geometry of the sphere, we can relate the radius of the top surface of the water (\( x \)) to the height of the water (\( h \)) using the equation of the sphere. The relationship is given by:
\[ R^2 = x^2 + (R - h)^2 \]
Substituting \( R = 10 \):
\[ 10^2 = x^2 + (10 - h)^2 \]

### Step 3: Substitute Known Values
When \( h = 4 \):
\[ 10^2 = x^2 + (10 - 4)^2 \]
This simplifies to:
\[ 100 = x^2 + 6^2 \]
\[ 100 = x^2 + 36 \]
\[ x^2 = 100 - 36 = 64 \]
Thus, \( x = 8 \) cm.

### Step 4: Differentiate with Respect to Time
Now, we differentiate the equation \( 10^2 = x^2 + (10 - h)^2 \) with respect to time \( t \):
\[ 0 = 2x \frac{dx}{dt} + 2(10 - h)(- \frac{dh}{dt}) \]
This simplifies to:
\[ 0 = 2x \frac{dx}{dt} - 2(10 - h) \frac{dh}{dt} \]

### Step 5: Solve for \( \frac{dx}{dt} \)
Rearranging gives:
\[ 2x \frac{dx}{dt} = 2(10 - h) \frac{dh}{dt} \]
Dividing both sides by 2:
\[ x \frac{dx}{dt} = (10 - h) \frac{dh}{dt} \]
Thus,
\[ \frac{dx}{dt} = \frac{(10 - h)}{x} \frac{dh}{dt} \]

### Step 6: Substitute Known Values
Now we substitute \( h = 4 \), \( \frac{dh}{dt} = -2 \), and \( x = 8 \):
\[ \frac{dx}{dt} = \frac{(10 - 4)}{8} \cdot (-2) \]
This simplifies to:
\[ \frac{dx}{dt} = \frac{6}{8} \cdot (-2) = \frac{3}{4} \cdot (-2) = -\frac{3}{2} \text{ cm/sec} \]

### Final Answer
The radius of the top surface of the water is decreasing at the rate of \( \frac{3}{2} \) cm/sec.