Answer :
Answer:
The function that represents the amount of air is [tex]\Delta V =\cfrac 43 \pi (3r^3+3r+1)[/tex]
Step-by-step explanation:
The amount of air here represents the difference between V(r) and V(r+1), so we can start working by finding an expression for the volume at r+1, and then subtract the original volume V(r).
Volume of balloon of radius r+1 inches.
We can replace r with r+1 on the formula and we get:
[tex]V(r+1)=\cfrac43 \pi (r+1)^3[/tex]
We can expand [tex](r+1)^3[/tex] since we will use it to simplify it later on.
So we will have first
[tex](r+1)^2 = (r+1)(r+1)\\(r+1)^2 =r^2+r+r+1\\(r+1)^2 = r^2+2r+1[/tex]
We can multiply that result by (r+1) to get [tex](r+1)^3[/tex]
[tex](r+1)^3= (r+1)^2 (r+1)\\(r+1)^3=(r^2+2r+1)(r+1)\\(r+1)^3= r^3+2r^2+r+r^2+2r+1\\(r+1)^3 =r^3+3r^2+3r+1[/tex]
Thus the volume equation at r+1 will be
[tex]V(r+1)=\cfrac 43 \pi (r^3+3r^2+3r+1)[/tex]
Finding the amount of air required to inflate from r to r+1
The amount required to inflate is the difference of volumes, so we have
[tex]V(r+1)-V(r)=\cfrac 43 \pi (r^3+3r^2+3r+1) \cfrac 43 \pi r^3[/tex]
Combinging both into one term by factor [tex]\cfrac 43 \pi[/tex] give us
[tex]V(r+1)-V(r)=\cfrac 43 \pi (r^3+3r^2+3r+1-r^3)[/tex]
Simplifying
[tex]V(r+1)-V(r)=\cfrac 43 \pi (3r^2+3r+1)[/tex]
And that function represents the amount required to inflate the balloon from r to r+1 inches.