(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 40 cm3s¡1?
(b) Find an expression for dy/dx
given that 3xy + ey = 3.
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} $.