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

(b) Find an expression fordy/dx

given that 3xy+ey= 3.

Printable View

- Jan 15th 2009, 05:10 PMFazrate 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*cm*3*s¡*1?

(b) Find an expression for*dy/dx*

given that 3*xy*+*ey*= 3.

- Jan 15th 2009, 05:16 PMMush
1) The volume of a sphere is $\displaystyle V = \frac{4}{3} \pi r^3 $. We know that the volume and radius are functions of time here, hence, using implicit differentiation:

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

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

Hence:

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

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

2) $\displaystyle 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!

$\displaystyle y(x+e) = 3 $

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

Then use the quotient rule!

Or

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

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

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

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

Rearrange and remember to substitute in y again from the fact that $\displaystyle y = \frac{3}{x+e} $. - Jan 15th 2009, 05:22 PMtrythisThanks & noted
Thank you