# Related Rates QuestionSuppose hat a droplet of mist is a perfct sphere and that, through condensation, the droplet picks up moisture at a rate porportional to its surface area. Show that under...

Related Rates Question

Suppose hat a droplet of mist is a perfct sphere and that, through condensation, the droplet picks up moisture at a rate porportional to its surface area. Show that under these circumstances the droplet's radius increases at a constant rate.

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You need to remember that the area of sphere is of `4pi*r^2` (r expresses the radius of sphere).

You know that the mass of droplet is changing at a rate proportional to its surface area, hence:

`(dm)/(dt) ` = `c*A` ( c expresses the proportional constant)

Hence, substituting`4pi*r^2` for A yields:

`(dm)/(dt) = c*(dm)/(dr)*(dr)/(dt)`

You need to express mass in terms of radius, hence you need to use density formula such that:

mass = density*volume => mass = `rho*4*pi*r^3/3` `(rho` expresses density of water)

Differentiating the density formula with respect to radius yields:

`(dm)/(dr) = 4*rho*pi*3*r^2/3`

`(dm)/(dr) = 4*rho*pi*r^2`

You need to substitute 4`*rho*pi*r^2 for (dm)/(dr)` in `c*(dm)/(dr)*(dr)/(dt)` such that:

`c*(dm)/(dr)*(dr)/(dt) =c*(4*rho*pi*r^2)*(dr)/(dt)`

You need to substitute to the left `4pi*r^2*c` for `c*(dm)/(dr)*(dr)/(dt)` such that:

`4pi*r^2*c =c*(4*rho*pi*r^2)*(dr)/(dt)`

You need to reduce 4`*pi*c*r^2` both sides such that:

`(dr)/(dt) = 1/rho`

**Hence, since `(dr)/(dt)` expresses the changing of droplet's radius, the last equation shows that droplet's radius increases at a constant rate.**