If x,y ∈ 2n when n ∈ I and 1+logₓy=log₂y, then the value of (x+y) is
Answers
Step-by-step explanation:
Answer:
Rate at which volume of the bubble is increasing = 72 π cm³/s
Step-by-step explanation:
Given:
Radius of an air bubble is increasing at the rate of 2 cm/s
To Find:
The rate at which the volume of the bubble is increasing when the radius of the air bubble is 3 cm.
Solution:
Let the radius of the air bubble be r cm and volume be V.
Here the air bubble is in the shape of a sphere.
Volume of a sphere = 4/3 × π × r³
Now the rate of volume change with respect to time is given by,
\sf \dfrac{dV}{dt} = \dfrac{d}{dt} (\dfrac{4}{3}\: \pi r^{3})
dt
dV
=
dt
d
(
3
4
πr
3
)
Using chain rule,
\sf \dfrac{dV}{dt} =\dfrac{d}{dr} (\dfrac{4}{3}\: \pi r^{3} ).\dfrac{dr}{dt}
dt
dV
=
dr
d
(
3
4
πr
3
).
dt
dr
Differentiating,
\sf \dfrac{dV}{dt} =\dfrac{4}{3}\: \pi \times 3r^{2} .\dfrac{dr}{dt}
dt
dV
=
3
4
π×3r
2
.
dt
dr
Now the rate of increase of radius of the air bubble is given as,
\sf \dfrac{dr}{dt} =2\:cm/s
dt
dr
=2cm/s
Also by given, the radius of the air bubble is 3 cm.
Substitute the data,
\sf \dfrac{dV}{dt} =\dfrac{4}{3}\: \pi \times 3\times 3^{2} \times 2
dt
dV
=
3
4
π×3×3
2
×2
Simplifying we get,
\sf \dfrac{dV}{dt} =4\times \pi \times 9\times 2
dt
dV
=4×π×9×2
⇒ 72 π cm³/s
Therefore the rate at which the volume of the air bubble is increasing is 72 π cm³/s.