Respuesta :
we are given
volume of sphere as
[tex]V=\frac{4}{3} \pi r^3[/tex]
[tex]r(t)=3t[/tex]
(a)
we can plug r into V
we get
[tex]V=\frac{4}{3} \pi (3t)^3[/tex]
[tex]V(t)=36\pi t^3[/tex]
(b)
we can use average rate of change formula
[tex]A=\frac{f(b)-f(a)}{b-a}[/tex]
here , we are given t=1 to t=4
so, we get formula as
[tex]A=\frac{V(4)-V(1)}{4-1}[/tex]
now, we can plug values
[tex]A=\frac{36 \pi (4)^3-36 \pi (1)^3}{4-1}[/tex]
[tex]A=756\pi[/tex].............Answer
The volume of the spherical hot air balloon is [tex]\rm 36\pi t^3[/tex].
The average rate of change of the volume with respect to t as t changes from t = 1 to t = 4 is [tex]756\pi[/tex].
Given
The volume of a spherical hot air balloon as its radius changes.
The radius is a function of time given by r(t) = 3t.
What is the average rate of change?
The Average Rate of Change function is defined as the average rate at which one quantity is changing with respect to something else changing.
The volume of the spherical hot air balloon is given by;
[tex]\rm Volume = \dfrac{4}{3}\pi r^3[/tex]
Where r is the radius of the balloon.
1. The volume of the spherical hot air balloon is;
[tex]\rm Volume = \dfrac{4}{3}\pi r^3\\\\r(t)=3t\\\\\rm Volume = \dfrac{4}{3}\pi (3t)^3\\\\\rm Volume = 4 \pi \times 9t^3\\ \\Volume = 36\pi t^3[/tex]
The volume of the spherical hot air balloon is [tex]\rm 36\pi t^3[/tex].
2. The average rate of change of the volume with respect to t as t changes from t = 1 to t = 4 is;
[tex]\rm Average \ rate =\dfrac{v(4)-v(1)}{4-1}\\\\Average \ rate =\dfrac{36\pi (4)^3-36\pi (1)^3}{3}\\\\Average \ rate =\dfrac{2304\pi -36\pi }{3}\\\\Average \ rate = \dfrac{2268\pi }{3}\\\\Average \ rate =756\pi[/tex]
The average rate of change of the volume with respect to t as t changes from t = 1 to t = 4 is [tex]756\pi[/tex].
To know more about the Average rate click the link given below.
https://brainly.com/question/5049895
