Water is increasing in a rigid circular cone with rate `1 cm^3//sec`, then find the rate of change of lateral surface of cone is (where hight and diameter of cone is 35 and 14 resp) at h=10cm

Options

A. `sqrt26/10`
B. `sqrt26/5`
C. `sqrt21/5`
D. 5

Answer

To solve the problem, we need to find the rate of change of the lateral surface area of a cone as water is being added to it. Let's break down the solution step by step.

### Given:
- Height of the cone \( H = 35 \) cm
- Diameter of the cone \( D = 14 \) cm, hence the radius \( R = \frac{D}{2} = 7 \) cm
- Volume of water being added \( \frac{dV}{dt} = 1 \) cm³/sec
- Height of water at the moment we are considering \( h = 10 \) cm

### Step 1: Find the relationship between height and radius
Since the cone is rigid and maintains its shape, the radius \( r \) of the water level at height \( h \) is proportional to \( h \). We can set up a ratio using similar triangles:

\[ \frac{r}{h} = \frac{R}{H} \implies \frac{r}{h} = \frac{7}{35} \implies r = \frac{h}{5} \]

### Step 2: Express the lateral surface area \( A \) of the cone
The formula for the lateral surface area of a cone is given by:

\[ A = \pi r l \]

where \( l \) is the slant height of the cone. The slant height can be found using the Pythagorean theorem:

\[ l = \sqrt{r^2 + h^2} \]

### Step 3: Substitute \( r \) in terms of \( h \)
From the relationship we found earlier, substitute \( r = \frac{h}{5} \):

\[ l = \sqrt{\left(\frac{h}{5}\right)^2 + h^2} = \sqrt{\frac{h^2}{25} + h^2} = \sqrt{\frac{h^2 + 25h^2}{25}} = \sqrt{\frac{26h^2}{25}} = \frac{h\sqrt{26}}{5} \]

### Step 4: Substitute \( r \) and \( l \) into the area formula
Now substitute \( r \) and \( l \) into the area formula:

\[ A = \pi \left(\frac{h}{5}\right) \left(\frac{h\sqrt{26}}{5}\right) = \frac{\pi h^2 \sqrt{26}}{25} \]

### Step 5: Differentiate \( A \) with respect to time \( t \)
To find the rate of change of the lateral surface area with respect to time, we differentiate \( A \):

\[ \frac{dA}{dt} = \frac{d}{dt}\left(\frac{\pi h^2 \sqrt{26}}{25}\right) = \frac{\pi \sqrt{26}}{25} \cdot 2h \frac{dh}{dt} \]

### Step 6: Find \( \frac{dh}{dt} \)
We know the volume of the cone is given by:

\[ V = \frac{1}{3} \pi r^2 h \]

Substituting \( r = \frac{h}{5} \):

\[ V = \frac{1}{3} \pi \left(\frac{h}{5}\right)^2 h = \frac{1}{3} \pi \frac{h^3}{25} = \frac{\pi h^3}{75} \]

Differentiating both sides with respect to \( t \):

\[ \frac{dV}{dt} = \frac{\pi}{75} \cdot 3h^2 \frac{dh}{dt} \]

Setting \( \frac{dV}{dt} = 1 \):

\[ 1 = \frac{\pi}{75} \cdot 3h^2 \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{75}{3\pi h^2} = \frac{25}{\pi h^2} \]

### Step 7: Substitute \( h = 10 \) cm into \( \frac{dh}{dt} \)
Now substituting \( h = 10 \):

\[ \frac{dh}{dt} = \frac{25}{\pi (10)^2} = \frac{25}{100\pi} = \frac{1}{4\pi} \]

### Step 8: Substitute \( h \) and \( \frac{dh}{dt} \) into \( \frac{dA}{dt} \)
Finally, substitute \( h = 10 \) and \( \frac{dh}{dt} = \frac{1}{4\pi} \) into \( \frac{dA}{dt} \):

\[ \frac{dA}{dt} = \frac{\pi \sqrt{26}}{25} \cdot 2(10) \cdot \frac{1}{4\pi} = \frac{\sqrt{26}}{25} \cdot 5 = \frac{\sqrt{26}}{5} \]

### Final Answer
The rate of change of the lateral surface area of the cone when the height of the water is 10 cm is:

\[ \frac{dA}{dt} = \frac{\sqrt{26}}{5} \text{ cm}^2/\text{sec} \]