Respuesta :
Answer:
1130.97 cm/min.
Step-by-step explanation:
The volume of a sphere is given by the following equation:
[tex]V = \frac{4\pi r^{3}}{3}[/tex]
We have to do the implicit differentitaiton of V in function of t. The are only two variables(V and r). So
[tex]\frac{dV}{dt} = \frac{4\pi}{3}3r^{2}\frac{dr}{dt}[/tex]
[tex]\frac{dV}{dt} = 4\pi r^{2} \frac{dr}{dt}[/tex]
We have to find [tex]\frac{dV}{dt}[/tex] when [tex]\frac{dr}{dt} = -0.4, r = 15[/tex]. So
[tex]\frac{dV}{dt} = 4\pi*(15)^{2}*(-0.4)[/tex]
[tex]\frac{dV}{dt} = -1130.97[/tex]
The negative means that the volume is decreasing.
The volume of the snowball is decreasing at the rate of 1130.97 cm/min when the radius is 15 cm.
So the answer is 1130.97 cm/min.
When the radius is 15 cm, the volume will decrease at a rate of -1,130.4 cm^3/min
At what rate the volume of the snowball is decreasing?
We know that for a sphere of radius R, the volume is:
V = (4/3)*3.14*R^3
The radius decreases at a rate of 0.4 cm/min, and we want to find the rate when the radius is 15cm, then we can write:
R = 15cm - (0.4 cm/min)*t
Then the volume will be:
V(t) = (4/3)*3.14*(15cm - (0.4 cm/min)*t)^3
To get the rate of change, we need to differentiate with respect to t, we will get:
V'(t) = 3*(4/3)*3.14*(15cm - (0.4 cm/min)*t)^2*(-0.4 cm/min)
The rate of change when the radius is 15 cm, is V'(0), so we get:
V'(0min) = 3*(4/3)*3.14*(15cm - (0.4 cm/min)*0min)^2*(-0.4 cm/min)
V'(0min)= 3*(4/3)*3.14*(15cm)^2*(-0.4 cm/min) = -1,130.4 cm^3/min
If you want to learn more about rates of change, you can read:
https://brainly.com/question/8728504
