# rate of change

• Jan 15th 2009, 06:10 PM
Faz
rate of change
(a) At what rate is the radius of a spherical balloon changing at the time that the radius is 10
cm, if air is being pumped in at a rate of 40 cm3s¡1?

(b) Find an expression for
dy/dx

given that 3
xy + ey = 3.

• Jan 15th 2009, 06:16 PM
Mush
Quote:

Originally Posted by Faz
(a) At what rate is the radius of a spherical balloon changing at the time that the radius is 10
cm, if air is being pumped in at a rate of 40 cm3s¡1?

(b) Find an expression for
dy/dx

given that 3
xy + ey = 3.

1) The volume of a sphere is $V = \frac{4}{3} \pi r^3$. We know that the volume and radius are functions of time here, hence, using implicit differentiation:

$\frac{dV}{dt} = \frac{4}{3} \pi \times \frac{d}{dt}(r^3)$

$\frac{dV}{dt} = \frac{4}{3} \pi \times 3r^3 \times \frac{dr}{dt}$

Hence:

$\frac{dr}{dt} = \frac{dV}{dt} \times \frac{1}{4\pi r^2}$

You know what $\frac{dV}{dt}$ is, and you know what the radius is at our point of interest. So plug them in!

2) $xy+ey = 3$

You could take out a factor of y on the LHS, and then move all other terms to the RHS, giving you a nice explicit function to differentiate. Or you could go about it using implicit differentiation!

$y(x+e) = 3$

$y = \frac{3}{x+e}$

Then use the quotient rule!

Or

$\frac{d}{dx}(yx) + \frac{d}{dx}(ye) = \frac{d}{dx}(3)$

$x\frac{dy}{dx} + y\frac{dx}{dx} + e\frac{d}{dx}(y) = 0$

$x\frac{dy}{dx} + y + e\frac{dy}{dx} = 0$

$\frac{dy}{dx}(x+e) + y = 0$

Rearrange and remember to substitute in y again from the fact that $y = \frac{3}{x+e}$.
• Jan 15th 2009, 06:22 PM
trythis
Thanks & noted
Thank you