Answer:
The volume of spherical snow ball is decreasing at a rate 1088.952 cubic cm per minute.
Step-by-step explanation:
We are given the following in the question:
The radius of a melting snowball is decreasing.
[tex]\dfrac{dr}{dt} = -0.3\text{ cm per min}[/tex]
Instant radius = 17 cm
Volume of spherical snow ball =
[tex]V=\dfrac{4}{3}\pi r^3[/tex]
Rate of change of volume =
[tex]=\dfrac{dV}{dt} = \dfrac{d}{dt}(\dfrac{4}{3}\pi r^3)\\\\\dfrac{dV}{dt} = 4\pi r^2\times \dfrac{dr}{dt}\\\\\dfrac{dV}{dt} = 4(3.14)(17)^2\times (-0.3)\\\\\text{Negative sign shows decrease in volume}\\\\\dfrac{dV}{dt} = -1088.952\text{ cubic cm per min}[/tex]
Thus, the volume of spherical snow ball is decreasing at a rate 1088.952 cubic cm per minute.
The volume of the snowball decreasing when the radius is 17 cm at the rate "1089.5 cm³/min".
According to the question,
Radius is decreasing at rate, [tex]\frac{dx}{dt}[/tex] = -0.3 cm/min or,
Radius, r = 17 cm
Let,
The volume be "V".
We know the formula,
→ V = [tex]\frac{4}{3}[/tex] πr³
By differentiating the terms,
[tex]\frac{dV}{dt}[/tex] = [tex]\frac{4}{3}[/tex] π.3r² [tex]\frac{dx}{dt}[/tex]
When,
V = 17 cm and [tex]\frac{dx}{dt}[/tex] = -0.3 cm/min the,
→ [tex]\frac{dv}{dt}[/tex] = 4 π (17)².(-0.3)
= 1089.5 cm³/min
Thus the above answer is correct.
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