**PLS HELP I DONT GET IT**

The instantaneous rate of change of the volume V of a sphere with respect to its radius r can be expressed as (dV) / (dr) = a(V / r). What is a? Recall that the volume formula for a sphere is V=(4 / 3)πr^3. Show all your work.

PLS HELP I DONT GET IT The instantaneous rate of change of the volume V of a sphere with respect to its radius r can be expressed as dV dr aV r What is a Recall class=

Respuesta :

Answer:

3

Step-by-step explanation:

V = ⁴/₃ π r³

Taking derivative of both sides with respect to r, we get:

dV/dr = 4π r²

Going back to the volume equation, if we divide both sides by r:

V/r = ⁴/₃ π r²

dV/dr is 3 times greater than V/r. Therefore, α = 3.

4π r² = 3 (⁴/₃ π r²)

dV/dr = 3 V/r

Answer:

a = 3

Step-by-step explanation:

The volume of a sphere is given by the formula V = (4/3)πr³.

To find dV/dr, we can differentiate both sides of this equation with respect to r:

[tex]\dfrac{\text{d}V}{\text{d}r}=3 \cdot \dfrac{4}{3} \pi r^{3-1}[/tex]

[tex]\dfrac{\text{d}V}{\text{d}r}=4 \pi r^{2}[/tex]

The instantaneous rate of change of the volume V of a sphere with respect to its radius r can be expressed as dV/dr = a(V/r). To find the value of a, we can substitute the expressions for dV/dr and V into the equation:

[tex]\begin{aligned}\dfrac{\text{d}V}{\text{d}r}&=a\:\dfrac{V}{r}\\\\4 \pi r^{2}&=a\:\dfrac{\dfrac{4}{3} \pi r^3}{r}\end{aligned}[/tex]

Solve for a:

[tex]\begin{aligned}4 \pi r^{2}\cdot r&=a\:\dfrac{\dfrac{4}{3} \pi r^3}{r}\cdot r\\\\4 \pi r^{3}&=a\:\dfrac{4}{3} \pi r^3\\\\3 \cdot 4\pi r^3&=a\;4\pi r^3\\\\3&=a\end{aligned}[/tex]

Therefore, the value of a is:

[tex]\huge\boxed{\boxed{a=3}}[/tex]

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