The surface area of a balloon is given by S(r) = 4(pie)r^2, where r is the radius of the balloon. If the radius is increasing with time t, as the balloon is being blown up, according to the formula r(t)=4/5t^3,t>=0, find the surface area S as a function of the time t.

r= 4/5 t^3

S=4PI (4/5 t^3)^2 and then you simplify that.

To find the surface area S as a function of time t, we need to substitute the given expression for r(t) into the surface area formula S(r).

Given: r(t) = (4/5)t^3, t >= 0

Surface area formula: S(r) = 4πr^2

Substituting r(t) into S(r):

S(t) = 4π(4/5t^3)^2
= 4π(16/25)t^6
= (64/25)πt^6

Therefore, the surface area S as a function of time t is S(t) = (64/25)πt^6.

To find the surface area S as a function of time t, we need to substitute the given expression for r(t) into the formula for S(r).

Given:
r(t) = (4/5)t^3, t >= 0

Substitute r(t) into the formula for S(r):
S(t) = 4πr^2

Replace r with r(t):
S(t) = 4π((4/5)t^3)^2

Now let's simplify this expression:

Step 1: Square the term inside the parentheses.
S(t) = 4π((16/25)t^6)

Step 2: Multiply the constant 4π with the expression (16/25)t^6.
S(t) = (64/25)πt^6

Therefore, the surface area S as a function of time t is given by:
S(t) = (64/25)πt^6

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