# Calculus on parametric curves.

## Curve sketching.

• In chapter 4, we have learnt how get information about a graph $$C:y=f(x)$$ via the derivatives of $$f(x)$$:
• For a point $$P(a,f(x=a))$$ on $$C$$, the derivative $$f'(x=a)$$ gives the slope of the tangent line of $$C$$. Correspondingly, an equation of the tangent line is $y-f(x=a)=f'(x=a)(x-a).$
• This also tells us that the function $$f(x)$$ is increasing when $$f'(x)>0$$, and $$f(x)$$ is decreasing, when $$f'(x)<0$$.
• The second derivative $$f''(x)$$ gives information about concavity.
• Recall that a curve is concave upward if all of its tangent lines are below the curve. A curve is concave downward if all of its tangent lines are above the curve.
• The graph $$C:y=f(x)$$ is concave upward where $$f''(x)>0$$, and it's concave downward where $$f''(x)<0$$.
• We can directly translate these statements to the case of a parametric curve $C:x=f(t),\quad y=g(t)$ using the chain rule.
• For the first derivative, we get $\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}=\frac{g'(t)}{f'(t)}.$
• Therefore, for a point $$P(f(t=a),g(t=a))$$ on $$C$$, an equation for the tangent line at $$P$$ is $y-g(t=a)=\frac{g'(t=a)}{f'(t=a)}(x-f(t=a)).$
• Note that the slope $$\frac{g'(t)}{f'(t)}$$ is calculated with respect to the $$(x,y)$$ coordinate system, and you need to interpret your results accordingly.
• For example, in the case of the standard parametrization of the unit circle: $x=\cos t,\quad y=\sin t,\quad0\le t\le2\pi$ we have $$\frac{g'(t)}{f'(t)}=-\cot t$$. This function is positive when $$\frac{\pi}{2}<t<\pi$$ or $$\frac{3\pi}{2}<t<2\pi$$. Note that those are the areas where increasing $$x$$ increases $$y$$.
• We can get a formula for the second derivative by using the chain rule again: $\frac{\mathrm d}{\mathrm dx}\left(\frac{\mathrm dy}{\mathrm dx}\right)=\frac{\frac{\mathrm d}{\mathrm dt}\left(\frac{\mathrm dy}{\mathrm dx}\right)}{\frac{\mathrm dx}{\mathrm dt}}.$
• In the case of $$f(t)=\cos t,\,g(t)=\sin t$$, we get $\frac{\mathrm d^2y}{\mathrm dx^2}=\frac{(-\cot t)'}{(\cos t)'}=-\frac{\csc^2t}{\sin t}=-\frac{1}{\sin^3t}.$ This is positive precisely when $$\pi<t<2\pi$$. Note that that's indeed where the circle is concave upward.
• Exercises. 10.2: 3, 5, 11, 17, 19, 27, 29

## Arc length

• Consider a graph $$C:y=F(x),\,a\le x\le b$$. Suppose that $$C$$ can also be described with the parametric equations $$x=f(t),\,y=g(t),\,\alpha\le t\le\beta$$, where $$f'(t)>0$$. Then we can use the substitution formula to rewrite the arc length formula: $L=\int_a^b\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx=\int_\alpha^\beta\sqrt{1+\left(\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}\right)^2}\frac{\mathrm dx}{\mathrm dt}\mathrm dt=\int_\alpha^\beta\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt.$
• Theorem. The same formula works for any parametric curve $$x=f(t),\,y=g(t),\,\alpha\le t\le\beta$$, even when it is not a graph.
• Example. Let's calculate the arc length function of a circle of radius $$r$$: $x=r\cos\theta,\,y=r\sin\theta,\,\theta\ge0.$
• We have $$f'(\theta)=-r\sin\theta$$ and $$g'(\theta)=r\cos\theta$$, therefore we get $s(\theta)=\int_0^\theta\sqrt{(r\sin t)^2+(r\cos t)^2}\,\mathrm dt=\int_0^\theta r\,\mathrm dt=r\theta.$
• Example. The curve traced out by a point $$P$$ on the circumference of a circle of radius $$r$$ as the circle rolls along a straight line is called a cycloid. We can parametrize it as follows.
• Let $$\theta$$ be the angle of $$P$$ in the circle with respect to the rightmost point.
• We have seen in the previous example that as $$P$$ travels an angle of $$\theta$$ on the circle, it covers a path length of $$r\theta$$. Therefore, at angle $$\theta$$, the centre of the circle is at $x_1=r\theta,\quad y_1=r.$
• On the circle, the point is moving in the clockwise direction. For some reason, it's customary to make it start at the bottom. This can be parametrized by $x_2=-r\sin\theta,\quad y_2=-r\cos\theta.$
• To get the parametrization of the cycloid, we can add the two functions: $x=x_1+x_2=r(\theta-\sin\theta),\quad y=r(1-\cos\theta)$
• To get at the arc length, we first compute the derivatives: $f'(\theta)=r(1-\cos\theta),\quad g'(\theta)=r\sin\theta$
• Let's see the path length over a full revolution: $L=\int_0^{2\pi}\sqrt{r^2(1-\cos\theta)^2+r^2\sin^2\theta)}\,\mathrm d\theta=r\int_0^{2\pi}\sqrt{2-2\cos\theta}\,\mathrm d\theta=r\int_0^{2\pi}\sqrt{4\sin^2(\theta/2)}\,\mathrm d\theta\stackrel{\sin(\theta/2)\ge0\text{ for }0\le\theta\le2\pi}{=}2r\int_0^{2\pi}\sin(\theta/2)\,\mathrm d\theta=8r.$
• Exercises. 10.2: 41, 43