A spherical balloon is being inflated at the rate of 35cc/min. The rate of increase in the surface area (in cm$$^2$$/min.) of the balloon when its diameter is 14 cm, is :

We are given that a spherical balloon is being inflated at a rate of 35 cc/min, meaning the volume is increasing at 35 cm³/min. We need to find the rate of increase of the surface area when the diameter is 14 cm. The diameter is 14 cm, so the radius $$ r = \frac{14}{2} = 7 $$ cm.

Recall the formulas for a sphere:

• Volume, $$ V = \frac{4}{3} \pi r^3 $$
• Surface area, $$ S = 4 \pi r^2 $$

We know $$ \frac{dV}{dt} = 35 $$ cm³/min. We need to find $$ \frac{dS}{dt} $$ when $$ r = 7 $$ cm.

Both volume and surface area depend on the radius $$ r $$. To relate $$ \frac{dV}{dt} $$ and $$ \frac{dS}{dt} $$, we use the chain rule through $$ \frac{dr}{dt} $$.

First, differentiate the volume formula with respect to time $$ t $$:

$$ V = \frac{4}{3} \pi r^3 $$

Differentiating both sides:

$$ \frac{dV}{dt} = \frac{4}{3} \pi \cdot 3 r^2 \cdot \frac{dr}{dt} $$

Simplify:

$$ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} $$

Now plug in $$ \frac{dV}{dt} = 35 $$ and $$ r = 7 $$:

$$ 35 = 4 \pi (7)^2 \frac{dr}{dt} $$

Calculate $$ (7)^2 = 49 $$:

$$ 35 = 4 \pi \cdot 49 \cdot \frac{dr}{dt} $$

$$ 35 = 196 \pi \frac{dr}{dt} $$

Solve for $$ \frac{dr}{dt} $$:

$$ \frac{dr}{dt} = \frac{35}{196 \pi} $$

Simplify the fraction by dividing numerator and denominator by 7:

$$ \frac{dr}{dt} = \frac{5}{28 \pi} \text{ cm/min} $$

Next, differentiate the surface area formula with respect to time $$ t $$:

$$ S = 4 \pi r^2 $$

Differentiating both sides:

$$ \frac{dS}{dt} = 4 \pi \cdot 2r \cdot \frac{dr}{dt} $$

Simplify:

$$ \frac{dS}{dt} = 8 \pi r \frac{dr}{dt} $$

Plug in $$ r = 7 $$ and $$ \frac{dr}{dt} = \frac{5}{28 \pi} $$:

$$ \frac{dS}{dt} = 8 \pi \cdot 7 \cdot \frac{5}{28 \pi} $$

Notice that $$ \pi $$ in the numerator and denominator cancel:

$$ \frac{dS}{dt} = 8 \cdot 7 \cdot \frac{5}{28} $$

Calculate step by step:

$$ 8 \cdot 7 = 56 $$

$$ 56 \cdot \frac{5}{28} = \frac{56 \cdot 5}{28} $$

Simplify $$ \frac{56}{28} = 2 $$:

$$ 2 \cdot 5 = 10 $$

So, $$ \frac{dS}{dt} = 10 $$ cm²/min.

Now, comparing with the options:

• A. 10
• B. $$ \sqrt{10} $$
• C. 100
• D. $$ 10\sqrt{10} $$

Hence, the correct answer is Option A.