Answer :
Answer:
The radius is growing at a rate of [tex]\frac{dr}{dt} = \frac{10}{4 \pi} = 0.795[/tex] ft per minute.
Step-by-step explanation:
From the information given we know that
[tex]\frac{dV}{dt} = 10[/tex]
And we know as well that .
[tex]V = \frac{4}{3} \pi r^3[/tex]
Since everything is changing with the time we can compute the implicit derivative and we would get that
[tex]\frac{dV}{dt} = 10 = 4\pi r^2 \frac{dr}{dt}[/tex]
We are told that we are looking for how fast is the radius growing at the instant when the radius has reached 1 ft, therefore [tex]r = 1.[/tex]
And when we solve for [tex]\frac{dr}{dt} = \frac{10}{4 \pi} = 0.795[/tex]
Rate of increase in the radius of the balloon will be 0.80 cubic feet per minute when the radius is 1 feet.
Volume of sphere:
- Volume (V) of a sphere is given by the expression,
[tex]V=\frac{4}{3}\pi r^3[/tex]
Here, [tex]r=[/tex] Radius of the sphere
If we have to calculate the increase in the volume, differentiate
the expression with respect to time,
[tex]\frac{d}{dt}(V)=\frac{d}{dt}(\frac{4}{3}\pi r^3)[/tex]
[tex]V'=\frac{4}{3}\pi \frac{d}{dt} [(r)^3][/tex]
[tex]=\frac{4}{3}\pi (3r^2)\frac{dr}{dt}[/tex]
[tex]=4\pi r^2(\frac{dr}{dt})[/tex]
Given in the question,
[tex]V'=10 \text{ cubic feet per minute}[/tex]
[tex]r=1\text{ feet}[/tex]
[tex]10=4\pi (1)^2(r')[/tex]
[tex]r'=\frac{10}{4\pi}[/tex]
[tex]=0.795[/tex]
[tex]r'\approx 0.80[/tex] cubic feet per minute
Therefore, rate of increase in radius of the balloon will be 0.80 cubic feet per minute when the radius is 1 feet.
Learn more about the differentiation here,
https://brainly.com/question/15278071?referrer=searchResults