A spherical weather balloon is being inflated. The radius of the balloon is increasing at the rate of 7 cm/s. Express the surfac

Question

Phoenix

3 months ago

15

A spherical weather balloon is being inflated. The radius of the balloon is increasing at the rate of 7 cm/s. Express the surfac

e area of the balloon as a function of time t (in seconds). (Let S(0) = 0.) . S(t) = cm^2?

Mathematics

2 answers:

Svet_ta [12.7K] · Answer 1 · 2026-03-03T06:36:26+03:00

Svet_ta [12.7K]3 months ago

7 0

R(t) = 6t

substituting into the formula 

V = (4/3) pi * r^3 

V(t) = (4/3) pi * (6t)^3 
 
V(t) = (4/3) pi * 216 t^3

Now multiply 4/3 by 216. 

V(t) = 288 * pi * t^3

zzz [12.3K] · Answer 2 · 2026-03-03T06:37:13+03:00

Answer:

$S(t)=196\pi t^2$

Step-by-step explanation:

It is given that

The radius of the balloon is increasing at a speed of 7 cm/s.

Let time equal t seconds.

In one second, the radius becomes 7 cm.

Therefore, in t seconds, the radius is R(t) = 7t.

We need to calculate the surface area of the balloon as a function of time t.

The area of a sphere can be represented as S(R)= $4\pi R^2$ .

Incorporate the value:

$S(R(t))=S(t)=4\pi (7t)^2=196\pi t^2$

Thus, $S(t)=196\pi t^2$