If the rate at which water vapor condenses onto a spherical raindrop is proportional to the surface area of the raindrop, show that the radius of the raindrop will increase at a constant rate?

Answer 1

Let us setup the following variables:

{

(r, "Radius of raindrop at time t","(cm)"), (S, "Surface area of raindrop at time t", "(cm"^3")"), (V, "Volume of raindrop at time t", "(cm"^2")"), (t, "time", "(sec)") :} #

The standard formula for Surface Area of a sphere, and the volume are:

# S = 4pir^2 \ \ #, and # \ \ V= 4/3 pi r^3 #
Differentiating wrt #r#, we have:
# (dV)/(dr)= 4 pi r^2 = S#

We are given that the rate at which water vapor condenses onto a spherical raindrop is proportional to the surface area of the raindrop, Thus:

# (dV)/(dt) prop S => (dV)/(dt) =k S \ \ #.....[A], say for some constant #k#

And os applying the chain rule, we have:

# (dV)/(dr) * (dr)/(dt) =k S #
# :. S (dr)/(dt) =k S #
# :. (dr)/(dt) =k #, a constant, QED
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Answer 2

The rate of change of the radius of the raindrop ((r)) is directly proportional to the rate at which water vapor condenses onto its surface ((dV/dt)). According to the given information, the rate of condensation ((dV/dt)) is proportional to the surface area of the raindrop ((4\pi r^2)). Therefore, the rate of change of the radius ((dr/dt)) is constant.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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