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a sphere is filled with air. if the volume of the sphere is increasing at a rate of 569 cubic inches per minute, what is the rate, in inches per minute, at which the radius of the sphere is changing when the radius is 4 inches? submit an exact answer. remember that the volume of a sphere is v

Sagot :

The rate, in inches per minute, at which the radius of the sphere is changing when the radius is 4 inches is 2.83

What is the Volume of a Sphere?

The capacity of a sphere is its volume. It is the area that the sphere occupies. Cubic measurements of a sphere's volume include m3, cm3, in3, etc. The sphere has a circular, three-dimensional form. Its form is determined by three axes: the x, y, and z axes. Sports like basketball and football are all instances of spheres with volume.

Since the cross-section of the sphere is a circle, the volume in this case is dependent on the diameter of the sphere's radius. The area or region of a sphere's outer surface is known as its surface area. We may use the following formula to get the volume of a sphere whose radius is "r":

Here, In the question, It is given that:

Radius = 4inch.

Rate of increase = 569 cubic inches;

So, using the Volume of Sphere, formula:

[tex]\begin{aligned}v=\frac{4}{3} \pi r^3 \\\frac{d v}{d t} &=\frac{d}{d t}\left[\frac{4}{3} \pi r^3\right] \\&=\frac{4 \pi}{3}\left(3 r^2\right) \frac{d r}{d t} \\\frac{d v}{d t} &=4 \pi r^2 \frac{d r}{d t}\end{aligned}[/tex]

[tex]\begin{aligned}&\frac{d r}{d t}=\frac{\frac{d v}{d t}}{4 \pi r^2}\\&\left.\frac{d r}{d t}=\frac{569}{(4 \pi)(4)^2}=\frac{d}{d t}_{r=4}^{d r^2}=569\right]\\&\frac{d v}{d t}=\frac{569}{64 \pi} \sim 2.83\end{aligned}[/tex]

Hence,  the rate, in inches per minute, at which the radius of the sphere is changing when the radius is 4 inches is 2.83

To learn more about Sphere, visit:

https://brainly.com/question/1122024

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