P4.5 - Differentiation

A curve need not be written as $y=f(x)$ before its gradient can be found. For an implicit equation, differentiate every term with respect to $x$ and attach $dy/dx$ whenever a function of $y$ is differentiated. For a parametric curve, compare how both coordinates change with the parameter.

For $x^2+xy+y^2=7$, differentiation gives $$2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0,$$ so $$\frac{dy}{dx}=-\frac{2x+y}{x+2y}.$$ At $(1,2)$ the tangent gradient is $-4/5$; the normal gradient is $5/4$.

For $x=t^2+1$ and $y=t^3-3t$, $$\frac{dy}{dx}=\frac{3t^2-3}{2t}.$$ At $t=2$, the point is $(5,2)$ and the gradient is $9/4$, so the tangent is $$y-2=\frac94(x-5).$$

At a point $(x_0,y_0)$, use $y-y_0=m(x-x_0)$. A finite non-zero tangent gradient $m$ gives normal gradient $-1/m$. Keep the point and gradient exact until the final line is simplified.

A differential equation states how a quantity changes, rather than giving the quantity directly. Build it by naming the changing variables, translating each rate statement with its sign and units, and differentiating the model relation with respect to the stated independent variable.

Suppose a tank has constant horizontal area $10\,\mathrm{m^2}$, water enters at $5\,\mathrm{m^3\,min^{-1}}$, and leaves at $0.2h\,\mathrm{m^3\,min^{-1}}$ when the depth is $h$ metres. Since $V=10h$, $$\frac{dV}{dt}=5-0.2h\quad\text{and}\quad\frac{dV}{dt}=10\frac{dh}{dt},$$ so the model is $10\,dh/dt=5-0.2h$.

For connected rates: (1) express all geometry in one changing variable; (2) differentiate before inserting the instant's values; (3) connect rates with the chain rule; (4) solve for the requested derivative; (5) report its sign and units.

For a cube of side $x$ cm, $S=6x^2$ and $V=x^3$. If its surface area decreases at $12\,\mathrm{cm^2\,s^{-1}}$, then $$-12=12x\frac{dx}{dt}.$$ At $x=3$, $dx/dt=-1/3\,\mathrm{cm\,s^{-1}}$, and $$\frac{dV}{dt}=3x^2\frac{dx}{dt}=-9\,\mathrm{cm^3\,s^{-1}}.$$