Modelling with Differentiation
- Up to now, we've had y in terms of x where \(\frac{dy}{dx}\) means the rate at which y changes in respect to x.
- But we can use similar gradient function notation for other physical quantities.
- E.G. A sewage container fills at a rate of \(20cm^3\) per second. To describe this, we can use \(\frac{dV}{dt}=20cm^3/s\)
INFO:
- Limitations: has to be a cuboid with a horizonal base with no top, must be made of \(54cm^2\) of sheet metal, must have two squares for sides.
- Surface area: \(54cm^2\)
- Side lengths of side squares: \(x\)
EQUATIONS:
- Volume: \(l\times w\times h\) = \(x^2+w\)
- Surface Area: \(2x^2+3xw \to2x^2+3xw=54\)
REARRANGING:
- \(3xw=54-2x^2\)
- \(w=\frac{18}{x}-\frac{2x}{3}\)
SUB BACK IN:
- \(V=x^2\times \left( \frac{18}{x}-\frac{2x}{3} \right)\)
- \(V=18x-\frac{2x^3}{3}\)
- \(V=18x-\frac{2}{3}x^3\)
FINDING THE OPTIMAL VALUE:
- Differentiate the function: \(18x-\frac{2x^3}{3} \to 18-2x^2\)
- \(18-2x^2=0\)
- \(x=^+_{-}3\)
- Sub back in to find the y value (or in this case, the volume)
- \(V=36\)
INFO:
- Limitations: has to be a cuboidwith volume of \(81cm^3\)
- Surface area: \(10x^2\)
- Side lengths of side squares: \(x\)
EQUATIONS:
- Volume: \(V=2x^2y\)
- Length: \(12x+4y\)
- \(y=\frac{81}{2x^2}\)
REARRANGING:
- \(L=12x+\frac{162}{x^2}\)
- \(\frac{dL}{dx}=12-324x^{-3}\)
- \(= 12-\frac{324}{x^3}\)
SET TO 0:
- \(12-\frac{324}{x^3}=0\)
- \(x^3=\frac{324}{12}=27\)
- \(x=3\)
FINDING THE OPTIMAL/MINIMUM VALUE:
- \(L_{\text{min}}=12(3)+\frac{162}{9}=54cm\)
- \(\frac{d^2L}{dx^2}=972x^{-4}\)
- When \(x=3\):
- \(\frac{d^2L}{dx^2}=\frac{972}{3^4}>0\therefore\) it is a minimum.
INFO:
- Perimeter: 40cm
- \(\pi r+2r+x=40\)
- \(A=\frac{\pi r^2}{2}+2rx\)
- \(x=40-\pi r-2r\)
- \(A=\frac{\pi r^2}{2}+2r(40-\pi r-2r)\)