Calculus 2

Derivatives of parametric curves

A seat on a spinning ride is placed by two formulas of time, one for how far east it is and one for how far north. How steep is its path at a given moment, and which way does the path bend, when the path loops back and crosses itself and is not the graph of any function?

You need

Second and higher derivativesCalculus 1 The chain ruleCalculus 1 Tangent lines and normal linesCalculus 1

Unlocks

Slope of a polar curveCalculus 2 Arc length and speed on parametric and polar curvesCalculus 2

Where you are 5 in Parametric, polar and vector valued functions

Calculus8

Parametric, polar and vector valued functions5

Area in polar coordinates

Derivatives of parametric curves

Slope of a polar curve Arc length and speed on parametric and polar curves Vector valued functions and motion in the plane

When you reach for it

Achola is studying the path of one seat on a teacup ride. A camera above the platform records where the seat is at each moment, and she fits two formulas to it: x(t) for the distance east of a mark on the platform and y(t) for the distance north, both in meters, with t in seconds. That is a curve given by a parameter, as in Parametric equations. The path loops, so it fails the vertical line test and is not the graph of any function y = f(x). Even when the parameter can be removed (Eliminating the parameter), the resulting equation is often not a function and is harder to differentiate than the two formulas themselves.

Yet steepness is a local idea. At a single moment, a small piece of the path near the seat usually is the graph of a function, and that piece has a slope. This lesson finds that slope straight from x(t) and y(t), writes tangent lines, finds where the path is level or upright, and then finds which way the path bends. Reach for it whenever a curve is described by a parameter and you need its slope, a tangent line, or its concavity at a given parameter value.

The procedure

Day 1: the slope and the tangent line

Suppose x(t) and y(t) have continuous derivatives, and at t = t_0 the horizontal rate \displaystyle \frac{dx}{dt} is not 0. Then \displaystyle \frac{dx}{dt} keeps one sign for t near t_0, so x is strictly increasing or strictly decreasing there and each x near x(t_0) comes from exactly one t. On that small piece the curve is the graph of a function, y = F(x), and y(t) = F\big(x(t)\big). That F is differentiable is the one-variable inverse function theorem, stated here without proof. The chain rule in the form of The chain rule now gives \displaystyle \frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}, and dividing by the nonzero \displaystyle \frac{dx}{dt}:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt}.

The slope is a ratio of two rates. If the seat moves 3 meters per second north while moving 2 meters per second east, its path rises \displaystyle \frac{3}{2} meters for each meter east. The tangent line at t_0 is then built exactly as in Tangent lines and normal lines: the point is \big(x(t_0), y(t_0)\big) and the slope is the ratio above at t_0.

Two special cases come straight from the ratio. Where \displaystyle \frac{dy}{dt} = 0 and \displaystyle \frac{dx}{dt} \ne 0, the slope is 0 and the tangent is horizontal. Where \displaystyle \frac{dx}{dt} = 0 and \displaystyle \frac{dy}{dt} \ne 0, the roles of x and y swap: near t_0 the curve is a graph x = G(y), and the same argument gives \displaystyle \frac{dx}{dy} = \frac{dx/dt}{dy/dt} = 0, so the tangent is the vertical line x = x(t_0). Where both rates are 0, the seat is momentarily at rest (compare Position, velocity and acceleration), the ratio reads \displaystyle \frac{0}{0}, and the formula says nothing; the limit of the ratio as t \to t_0 has to be taken instead.

One more consequence: a slope belongs to a parameter value, not to a point. A curve that passes through the same point at two different times can have a different tangent line on each pass.

Day 2: the second derivative and concavity

The slope \displaystyle m(t) = \frac{dy}{dx} is itself a function of t. The second derivative \displaystyle \frac{d^2y}{dx^2} is the rate of change of m with respect to x (Second and higher derivatives), and the same chain rule argument, applied to m in place of y, gives, wherever \displaystyle \frac{dx}{dt} \ne 0,

\begin{aligned} \frac{d^2y}{dx^2} &= \frac{d}{dx}\left(\frac{dy}{dx}\right) \\ &= \frac{dm/dt}{dx/dt}. \end{aligned}

Differentiate the slope with respect to t, then divide by \displaystyle \frac{dx}{dt} once more. Writing \displaystyle m = \frac{y'}{x'} (primes for t derivatives) and using the quotient rule gives a closed form that is useful for checking:

\frac{d^2y}{dx^2} = \frac{x'(t)\,y''(t) - y'(t)\,x''(t)}{\big(x'(t)\big)^3}.

Since near t_0 the curve is the graph of y = F(x) and this quantity is F''(x), its sign means what it means for any graph (Concavity and inflection points): positive, the path bends toward increasing y (concave up); negative, it bends toward decreasing y (concave down). This describes the shape of the path, whichever way the seat travels along it.

Worked examples

Example 1

Day 1: Achola’s teacup at one second

Over one stretch of the ride, Achola’s seat follows

x = 5 - t^2, \qquad y = t^3 - 4t

for -2.4 \le t \le 2.4. Find the slope of the path and the tangent line at t = 1, and say which way the seat is moving.

Solution

The seat is at x(1) = 4, y(1) = 1 - 4 = -3, the point (4, -3). The rates are \displaystyle \frac{dx}{dt} = -2t and \displaystyle \frac{dy}{dt} = 3t^2 - 4, so at t = 1 they are -2 and -1. The horizontal rate is not 0, so the ratio formula applies:

\begin{aligned} \frac{dy}{dx} &= \frac{3t^2 - 4}{-2t}, \\ \left.\frac{dy}{dx}\right|_{t=1} &= \frac{-1}{-2} = \frac{1}{2}. \end{aligned}

The tangent line through (4, -3) with slope \displaystyle \frac{1}{2} is \displaystyle y + 3 = \frac{1}{2}(x - 4), that is, \displaystyle y = \frac{1}{2}x - 5. Both rates are negative, so the seat is moving west and south at once; a line that you travel down and to the left rises to the right, which is why the slope is positive. Check: at x = 4 the line gives 2 - 5 = -3.

Example 2

Day 1: where the teacup’s path is level or upright

For the same path, find every point with a horizontal tangent and every point with a vertical tangent.

Solution

A horizontal tangent needs \displaystyle \frac{dy}{dt} = 3t^2 - 4 = 0, so \displaystyle t = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}, and there \displaystyle \frac{dx}{dt} = -2t \ne 0, so both qualify. With \displaystyle t^2 = \frac{4}{3}, \displaystyle x = 5 - \frac{4}{3} = \frac{11}{3} and \displaystyle y = t(t^2 - 4) = -\frac{8}{3}t. So \displaystyle t = \frac{2}{\sqrt{3}} gives \displaystyle \left(\frac{11}{3}, -\frac{16\sqrt{3}}{9}\right) and \displaystyle t = -\frac{2}{\sqrt{3}} gives \displaystyle \left(\frac{11}{3}, \frac{16\sqrt{3}}{9}\right), the bottom and top of the loop, about 3.08 meters below and above the mark’s level.

A vertical tangent needs \displaystyle \frac{dx}{dt} = -2t = 0, so t = 0, where \displaystyle \frac{dy}{dt} = -4 \ne 0. The tangent is the vertical line x = 5 at (5, 0). This fits the formula x = 5 - t^2 \le 5: the seat reaches its farthest point east at t = 0, moving straight south at 4 meters per second.

Example 3

Day 1: one spot, two tangent lines

Show that Achola’s path passes through one point twice, and find the tangent line on each pass.

Solution

Two times s \ne t give the same point when 5 - s^2 = 5 - t^2 and s^3 - 4s = t^3 - 4t. The first equation forces s = -t. Then y(-t) = -(t^3 - 4t) must equal t^3 - 4t, so t^3 - 4t = 0, giving t = 0 (rejected, since then s = t) or t = \pm 2. The point is x = 5 - 4 = 1, y = 0: the seat is at (1, 0) at t = -2 and again at t = 2.

From the slope formula of Example 1,

\begin{aligned} \left.\frac{dy}{dx}\right|_{t=-2} &= \frac{8}{4} = 2, \\ \left.\frac{dy}{dx}\right|_{t=2} &= \frac{8}{-4} = -2. \end{aligned}

On the first pass the tangent is y = 2(x - 1) = 2x - 2, and the seat moves east and north (\displaystyle \frac{dx}{dt} = 4, \displaystyle \frac{dy}{dt} = 8). On the second pass the tangent is y = -2x + 2, and the seat moves west and north. The point (1, 0) has no single slope; each time value does.

Figure 1Achola’s teacup path x = 5 - t^2, y = t^3 - 4t. The seat passes (1, 0) twice, at t = -2 and t = 2, with a different tangent line each time. The path is level at \displaystyle t = \mp\frac{2}{\sqrt{3}} and upright at t = 0.

Example 4

Day 1: Okello’s spinning car and its three cusps

On Okello’s ride, a disk of radius 1 meter rolls around the inside of a fixed ring of radius 3 meters, and a seat on the rim of the disk follows

\begin{aligned} x &= 2\cos t + \cos 2t, \\ y &= 2\sin t - \sin 2t, \end{aligned}

for 0 \le t \le 2\pi. Find the tangent line at \displaystyle t = \frac{\pi}{2}. Then find every t where \displaystyle \frac{dx}{dt} or \displaystyle \frac{dy}{dt} is 0, and describe the curve there.

Solution

The rates are \displaystyle \frac{dx}{dt} = -2\sin t - 2\sin 2t and \displaystyle \frac{dy}{dt} = 2\cos t - 2\cos 2t. The sum-to-product identities \displaystyle \sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} and \displaystyle \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}, with A = t, B = 2t, factor them:

\begin{aligned} \frac{dx}{dt} &= -4\sin\tfrac{3t}{2}\cos\tfrac{t}{2}, \\ \frac{dy}{dt} &= 4\sin\tfrac{3t}{2}\sin\tfrac{t}{2}. \end{aligned}

At \displaystyle t = \frac{\pi}{2} the seat is at x = 0 + \cos\pi = -1, y = 2 - \sin\pi = 2, and the rates are -2\cdot 1 - 2\cdot 0 = -2 and 2\cdot 0 - 2\cos\pi = 2. The slope is \displaystyle \frac{2}{-2} = -1 and the tangent line is y - 2 = -(x + 1), that is, y = -x + 1.

On 0 \le t < 2\pi (t = 2\pi repeats t = 0), \displaystyle \sin\frac{3t}{2} = 0 at \displaystyle t = 0, \frac{2\pi}{3}, \frac{4\pi}{3}, \displaystyle \cos\frac{t}{2} = 0 at t = \pi, and \displaystyle \sin\frac{t}{2} = 0 only at t = 0. So \displaystyle \frac{dx}{dt} = 0 at \displaystyle t = 0, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, and \displaystyle \frac{dy}{dt} = 0 at \displaystyle t = 0, \frac{2\pi}{3}, \frac{4\pi}{3}.

At t = \pi only the horizontal rate is 0, while \displaystyle \frac{dy}{dt} = 4(-1)(1) = -4: a vertical tangent at (-1, 0), the seat moving straight south. There is no horizontal tangent anywhere, because every zero of \displaystyle \frac{dy}{dt} is also a zero of \displaystyle \frac{dx}{dt}.

At \displaystyle t = 0, \frac{2\pi}{3}, \frac{4\pi}{3} both rates are 0 and the seat stops. Wherever the common factor \displaystyle \sin\frac{3t}{2} is not 0, it cancels:

\begin{aligned} \frac{dy}{dx} &= \frac{4\sin\frac{3t}{2}\sin\frac{t}{2}}{-4\sin\frac{3t}{2}\cos\frac{t}{2}} \\ &= -\tan\tfrac{t}{2}. \end{aligned}

As t \to 0, \displaystyle t \to \frac{2\pi}{3} and \displaystyle t \to \frac{4\pi}{3}, the slope tends to 0, \displaystyle -\tan\frac{\pi}{3} = -\sqrt{3} and \displaystyle -\tan\frac{2\pi}{3} = \sqrt{3}, at the points (3, 0), \displaystyle \left(-\frac{3}{2}, \frac{3\sqrt{3}}{2}\right) and \displaystyle \left(-\frac{3}{2}, -\frac{3\sqrt{3}}{2}\right). So each stopping point has a limiting tangent direction, yet the curve is not smooth there. Near t = 0, \displaystyle \frac{dx}{dt} is positive just before 0 and negative just after, while \displaystyle \frac{dy}{dt} \ge 0 on both sides: the seat comes in moving east along the lower arc, stops at (3, 0), and leaves moving west along the upper arc. Both arcs touch the line y = 0 there, one from below and one from above, which makes a sharp point, a cusp, with tangent line y = 0. The other two stopping points are cusps in the same way.

Figure 2Okello’s car x = 2\cos t + \cos 2t, y = 2\sin t - \sin 2t, traced counterclockwise. The seat stops at each of the three cusps, where \displaystyle \frac{dx}{dt} and \displaystyle \frac{dy}{dt} are both 0.

Example 5

Day 2: which way Achola’s loop bends

Find \displaystyle \frac{d^2y}{dx^2} for Achola’s path x = 5 - t^2, y = t^3 - 4t, and decide where the path is concave up and concave down. What happens at the point (1, 0) found in Example 3?

Solution

Write the slope from Example 1 as \displaystyle m(t) = \frac{3t^2 - 4}{-2t} = -\frac{3t}{2} + \frac{2}{t} and differentiate it with respect to t: \displaystyle \frac{dm}{dt} = -\frac{3}{2} - \frac{2}{t^2} = -\frac{3t^2 + 4}{2t^2}. Divide by \displaystyle \frac{dx}{dt} = -2t:

\begin{aligned} \frac{d^2y}{dx^2} &= \frac{-\frac{3t^2 + 4}{2t^2}}{-2t} \\ &= \frac{3t^2 + 4}{4t^3}, \quad t \ne 0. \end{aligned}

The closed form agrees: x' = -2t, x'' = -2, y' = 3t^2 - 4, y'' = 6t, so x'y'' - y'x'' = -12t^2 + 6t^2 - 8 = -6t^2 - 8, and dividing by (x')^3 = -8t^3 gives \displaystyle \frac{3t^2 + 4}{4t^3}.

The numerator is always positive, so the sign is the sign of t. For t > 0 (from (5, 0) along the bottom of the loop, through (1, 0) and up the tail beyond) the path is concave up; at t = 1 the value is \displaystyle \frac{7}{4}, so near (4, -3) the path lies above the tangent line \displaystyle y = \frac{1}{2}x - 5 of Example 1. For t < 0 (the lower tail and the top of the loop) it is concave down. At t = 0 the tangent is vertical and \displaystyle \frac{d^2y}{dx^2} does not exist.

At (1, 0) the two passes bend opposite ways: \displaystyle \frac{d^2y}{dx^2} = \frac{16}{32} = \frac{1}{2} at t = 2 and \displaystyle -\frac{1}{2} at t = -2. Like the slope, the concavity belongs to the parameter value.

Example 6

Day 2: the bend along Okello’s three arcs

Find \displaystyle \frac{d^2y}{dx^2} for Okello’s curve of Example 4, evaluate it at \displaystyle t = \frac{\pi}{2}, and find the concavity on each piece between the special values of t.

Solution

Away from the cusps, \displaystyle \frac{dy}{dx} = -\tan\frac{t}{2}, whose t derivative is \displaystyle -\frac{1}{2}\sec^2\frac{t}{2}. Dividing by \displaystyle \frac{dx}{dt} = -4\sin\frac{3t}{2}\cos\frac{t}{2}:

\begin{aligned} \frac{d^2y}{dx^2} &= \frac{-\frac{1}{2}\sec^2\frac{t}{2}}{-4\sin\frac{3t}{2}\cos\frac{t}{2}} \\ &= \frac{1}{8\sin\frac{3t}{2}\cos^3\frac{t}{2}}. \end{aligned}

At \displaystyle t = \frac{\pi}{2}: \displaystyle \sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2} and \displaystyle \cos^3\frac{\pi}{4} = \frac{\sqrt{2}}{4}, whose product is \displaystyle \frac{1}{4}, so \displaystyle \frac{d^2y}{dx^2} = \frac{1}{2}: concave up at (-1, 2).

The sign is the sign of \displaystyle \sin\frac{3t}{2}\cos^3\frac{t}{2}, and \displaystyle \cos^3\frac{t}{2} has the sign of \displaystyle \cos\frac{t}{2}.

t	\displaystyle \sin\frac{3t}{2}	\displaystyle \cos\frac{t}{2}	concave
\displaystyle \left(0, \frac{2\pi}{3}\right)	+	+	up
\displaystyle \left(\frac{2\pi}{3}, \pi\right)	-	+	down
\displaystyle \left(\pi, \frac{4\pi}{3}\right)	-	-	up
\displaystyle \left(\frac{4\pi}{3}, 2\pi\right)	+	-	down

So the upper arc is concave up, the left arc is concave down above the x-axis and concave up below it, and the lower arc is concave down. Every arc bows in toward the center, as the figure shows. The left arc switches concavity at (-1, 0) only because its tangent turns vertical there: seen as a graph x = G(y), that arc bends one way all along.

Boundaries

The ratio needs x(t) and y(t) differentiable at t_0. The curve x = t, y = \lvert t \rvert has a corner at t = 0 and no tangent line there.
Where \displaystyle \frac{dx}{dt} = 0 and \displaystyle \frac{dy}{dt} \ne 0, there is no slope as a number: the tangent is the vertical line x = x(t_0).
Where both rates are 0, the ratio is \displaystyle \frac{0}{0} and decides nothing. The limit of \displaystyle \frac{dy/dt}{dx/dt} as t \to t_0 may give a tangent direction at a cusp (Example 4), at a smooth point, or where the seat stops and turns back along its own path; and the limit may fail to exist.
A slope and a concavity belong to a value of t. A point the curve passes more than once may carry several of each.
\displaystyle \frac{d^2y}{dx^2} is not \displaystyle \frac{d^2y/dt^2}{d^2x/dt^2}. For Achola’s path at t = 1 that ratio is \displaystyle \frac{6}{-2} = -3, while the true value is \displaystyle \frac{7}{4}. The second derivative of y with respect to x describes the shape of the path, and the shape does not change if the seat runs the same path at another pace; the ratio of second t derivatives does change.
\displaystyle \frac{d^2y}{dx^2} also needs \displaystyle \frac{dx}{dt} \ne 0 and x, y twice differentiable; at a vertical tangent it does not exist.

Where this goes next

The two rates \displaystyle \frac{dx}{dt} and \displaystyle \frac{dy}{dt} give more than slope. Together they give how fast the seat travels and how far it goes, which is Arc length and speed on parametric and polar curves. A polar curve r = f(\theta) is a parametric curve with x = f(\theta)\cos\theta and y = f(\theta)\sin\theta, so its slope comes from the same ratio, in Slope of a polar curve. And the pair \displaystyle \left(\frac{dx}{dt}, \frac{dy}{dt}\right) is the velocity vector of Vector valued functions and motion in the plane: the tangent line points along it, which is why the tangent direction is in doubt exactly where the velocity is zero.

Exercises

Day 1: slopes and tangent lines

For x = t^2 + t, y = t^3 + 1, find \displaystyle \frac{dy}{dx} at t = 2 and the tangent line there.
For x = 3\sin t, y = 2\cos 2t, find the tangent line at \displaystyle t = \frac{\pi}{6}.
For x = t + \ln t, y = t^2 - 3t with t > 0, find the tangent line at t = 1.

Day 1: horizontal and vertical tangents

Find every point of x = t^2 - 6t, y = t^3 - 12t with a horizontal tangent and every point with a vertical tangent.
Shun’s ride carries a seat along x = 2\cos t, y = \sin t + \cos t for 0 \le t < 2\pi, in meters. Find the points where the path is level and where it is upright.
Find the horizontal and vertical tangents of x = t^3 - 3t^2, y = t^2 - 2t.

Day 1: crossings and stopping points

Show that x = t^2 + 1, y = t^3 - 5t passes through exactly one point twice, and find both tangent lines there.
The curve x = 2\sin 2t, y = 3\sin t, 0 \le t < 2\pi, passes through the origin twice. Find both tangent lines there.
For x = t^3, y = t^6 + 2t^3, both rates are 0 at t = 0. Find the limit of \displaystyle \frac{dy}{dx} as t \to 0 and the tangent line at the origin, and check the result by eliminating the parameter.

Day 2: second derivatives and concavity

For x = t^2 + t, y = t^3 + 1 (Exercise 1), find \displaystyle \frac{d^2y}{dx^2} as a function of t and its value at t = 2. Is the curve concave up or down there?
For x = e^t, y = t^2, find \displaystyle \frac{d^2y}{dx^2}, the values of t where the curve is concave up and concave down, and the point where the concavity changes.
For x = \ln t, y = t^3 with t > 0, find \displaystyle \frac{d^2y}{dx^2} and confirm it by eliminating the parameter.
For Shun’s path x = 2\cos t, y = \sin t + \cos t (Exercise 5), find \displaystyle \frac{d^2y}{dx^2}, its value at \displaystyle t = \frac{\pi}{2}, and the concavity on 0 < t < \pi and on \pi < t < 2\pi.
For x = t^2 - 6t, y = t^3 - 12t (Exercise 4), find \displaystyle \frac{d^2y}{dx^2} and the intervals of t on which the curve is concave up.
On Okello’s curve x = 2\cos t + \cos 2t, y = 2\sin t - \sin 2t, find the tangent line and \displaystyle \frac{d^2y}{dx^2} at \displaystyle t = \frac{\pi}{3}.

Reasoning

Someone claims that \displaystyle \frac{d^2y}{dx^2} = \frac{d^2y/dt^2}{d^2x/dt^2}. Test the claim on x = t^3 + t, y = t^2 at t = 1.
Explain why a point where \displaystyle \frac{dx}{dt} = 0 and \displaystyle \frac{dy}{dt} \ne 0 has a vertical tangent line.
Run a curve backward: replace x(t), y(t) by X(s) = x(-s), Y(s) = y(-s). Show that at each point the new \displaystyle \frac{dy}{dx} and \displaystyle \frac{d^2y}{dx^2} agree with the old ones (at s = -t), and explain why this had to happen.
For x = t^2, y = t^2 + t^4, show that both rates are 0 at t = 0 and find the limit of \displaystyle \frac{dy}{dx} there. Then describe the seat’s motion near t = 0, and explain why a limiting slope does not mean the seat passes through the point.
Show that for x = t, y = f(t), the formulas for \displaystyle \frac{dy}{dx} and \displaystyle \frac{d^2y}{dx^2} give f'(x) and f''(x).

Answers

\displaystyle \frac{dy}{dx} = \frac{12}{5}; tangent \displaystyle y = \frac{12}{5}x - \frac{27}{5}. The point is (6, 9); \displaystyle \frac{dx}{dt} = 2t + 1 = 5, \displaystyle \frac{dy}{dt} = 3t^2 = 12, and \displaystyle 9 - \frac{12}{5}\cdot 6 = -\frac{27}{5}.
\displaystyle y = -\frac{4}{3}x + 3. The point is \displaystyle \left(\frac{3}{2}, 1\right); \displaystyle \frac{dx}{dt} = 3\cos t = \frac{3\sqrt{3}}{2} and \displaystyle \frac{dy}{dt} = -4\sin 2t = -2\sqrt{3}, so the slope is \displaystyle -\frac{4}{3}.
\displaystyle y = -\frac{1}{2}x - \frac{3}{2}. The point is (1, -2); \displaystyle \frac{dx}{dt} = 1 + \frac{1}{t} = 2, \displaystyle \frac{dy}{dt} = 2t - 3 = -1, slope \displaystyle -\frac{1}{2}.
Horizontal at (-8, -16) and (16, 16); vertical at (-9, -9). \displaystyle \frac{dy}{dt} = 3t^2 - 12 = 0 at t = \pm 2, where \displaystyle \frac{dx}{dt} = 2t - 6 \ne 0; \displaystyle \frac{dx}{dt} = 0 at t = 3, where \displaystyle \frac{dy}{dt} = 15 \ne 0.
Level at \displaystyle (\sqrt{2}, \sqrt{2}) and \displaystyle (-\sqrt{2}, -\sqrt{2}); upright at (2, 1) and (-2, -1). \displaystyle \frac{dy}{dt} = \cos t - \sin t = 0 at \displaystyle t = \frac{\pi}{4}, \frac{5\pi}{4}, where \displaystyle \frac{dx}{dt} = -2\sin t \ne 0; \displaystyle \frac{dx}{dt} = 0 at t = 0, \pi, where \displaystyle \frac{dy}{dt} = \pm 1.
Horizontal at (-2, -1); vertical at (0, 0) and (-4, 0). \displaystyle \frac{dy}{dt} = 2t - 2 = 0 at t = 1, where \displaystyle \frac{dx}{dt} = -3; \displaystyle \frac{dx}{dt} = 3t(t - 2) = 0 at t = 0, 2, where \displaystyle \frac{dy}{dt} = -2 and 2.
The point is (6, 0), at \displaystyle t = \pm\sqrt{5}; tangents \displaystyle y = \sqrt{5}(x - 6) and \displaystyle y = -\sqrt{5}(x - 6). Equal x values with s \ne t force s = -t, and then y(-t) = -y(t) = y(t) forces t^3 - 5t = 0, so \displaystyle t = \pm\sqrt{5}. The slope \displaystyle \frac{3t^2 - 5}{2t} is \displaystyle \frac{10}{2\sqrt{5}} = \sqrt{5} at \displaystyle t = \sqrt{5} and \displaystyle -\sqrt{5} at \displaystyle t = -\sqrt{5}.
\displaystyle y = \frac{3}{4}x and \displaystyle y = -\frac{3}{4}x. The origin comes at t = 0 and t = \pi; \displaystyle \frac{dx}{dt} = 4\cos 2t = 4 at both, and \displaystyle \frac{dy}{dt} = 3\cos t is 3 and -3.
The limit is 2 and the tangent line is y = 2x. For t \ne 0, \displaystyle \frac{dy}{dx} = \frac{6t^5 + 6t^2}{3t^2} = 2t^3 + 2 \to 2. Since t^3 = x, the curve is y = x^2 + 2x, whose slope at x = 0 is 2: a smooth point.
\displaystyle \frac{d^2y}{dx^2} = \frac{6t(t + 1)}{(2t + 1)^3}, equal to \displaystyle \frac{36}{125} at t = 2: concave up. The slope is \displaystyle \frac{3t^2}{2t + 1}, with t derivative \displaystyle \frac{6t^2 + 6t}{(2t + 1)^2}; divide by 2t + 1.
\displaystyle \frac{d^2y}{dx^2} = 2(1 - t)e^{-2t}; concave up for t < 1, concave down for t > 1, changing at (e, 1). The slope is \displaystyle \frac{2t}{e^t} = 2te^{-t}, with t derivative 2(1 - t)e^{-t}; divide by e^t.
\displaystyle \frac{d^2y}{dx^2} = 9t^3. The slope is \displaystyle \frac{3t^2}{1/t} = 3t^3, with t derivative 9t^2, divided by \displaystyle \frac{1}{t}. Eliminating, t = e^x and y = e^{3x}, whose second derivative 9e^{3x} is 9t^3.
\displaystyle \frac{d^2y}{dx^2} = -\frac{1}{4\sin^3 t}, equal to \displaystyle -\frac{1}{4} at \displaystyle t = \frac{\pi}{2}; concave down on 0 < t < \pi, concave up on \pi < t < 2\pi. The slope is \displaystyle \frac{\cos t - \sin t}{-2\sin t} = \frac{1 - \cot t}{2}, with t derivative \displaystyle \frac{1}{2}\csc^2 t; divide by -2\sin t.
\displaystyle \frac{d^2y}{dx^2} = \frac{3(t^2 - 6t + 4)}{4(t - 3)^3}; concave up for \displaystyle 3 - \sqrt{5} < t < 3 and for \displaystyle t > 3 + \sqrt{5} (concave down for \displaystyle t < 3 - \sqrt{5} and \displaystyle 3 < t < 3 + \sqrt{5}). The slope \displaystyle \frac{3t^2 - 12}{2t - 6} has t derivative \displaystyle \frac{3(t^2 - 6t + 4)}{2(t - 3)^2}; divide by 2(t - 3). The numerator is negative only between its roots \displaystyle 3 \pm \sqrt{5}, and the denominator has the sign of t - 3.
Tangent \displaystyle y = -\frac{\sqrt{3}}{3}x + \frac{2\sqrt{3}}{3}; \displaystyle \frac{d^2y}{dx^2} = \frac{\sqrt{3}}{9}, concave up. The point is \displaystyle \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right); from Examples 4 and 6, the slope is \displaystyle -\tan\frac{\pi}{6} = -\frac{\sqrt{3}}{3} and \displaystyle \frac{d^2y}{dx^2} = \frac{1}{8\sin\frac{\pi}{2}\cos^3\frac{\pi}{6}} = \frac{1}{3\sqrt{3}}.
The claim is false: the true value is \displaystyle -\frac{1}{16}, the claimed one \displaystyle \frac{1}{3}. The slope is \displaystyle \frac{2t}{3t^2 + 1}, with t derivative \displaystyle \frac{2 - 6t^2}{(3t^2 + 1)^2} = -\frac{1}{4} at t = 1; dividing by \displaystyle \frac{dx}{dt} = 4 gives \displaystyle -\frac{1}{16}. The ratio of second derivatives is \displaystyle \frac{2}{6t} = \frac{1}{3}, which even has the wrong sign.
Since \displaystyle \frac{dy}{dt} \ne 0 (and is continuous), y is strictly monotone near t_0, so the curve is locally a graph x = G(y), and the chain rule gives \displaystyle \frac{dx}{dy} = \frac{dx/dt}{dy/dt} = 0. A tangent along which x does not change to first order is the vertical line x = x(t_0).
\displaystyle \frac{dY}{ds} = -y'(-s) and \displaystyle \frac{dX}{ds} = -x'(-s), so the slope at s is \displaystyle \frac{y'(t)}{x'(t)} with t = -s, unchanged. The new slope as a function of s is m(-s), whose s derivative -m'(-s) divided by -x'(-s) is again \displaystyle \frac{m'(t)}{x'(t)}. It had to happen because both quantities describe the shape of the path near the point, and running the same path backward does not change its shape.
\displaystyle \frac{dx}{dt} = 2t and \displaystyle \frac{dy}{dt} = 2t + 4t^3 are both 0 at t = 0; for t \ne 0, \displaystyle \frac{dy}{dx} = 1 + 2t^2 \to 1. Since x = t^2 \ge 0, the seat is on y = x + x^2 with x \ge 0 at all times: it slides down to the origin, stops at t = 0, and goes back up the same arc. The line y = x is tangent to the path at its end point, but the seat never moves beyond it; a limiting slope says which way the path points, not that the motion continues through.
With x = t, \displaystyle \frac{dx}{dt} = 1, so \displaystyle \frac{dy}{dx} = \frac{f'(t)}{1} = f'(x) and \displaystyle \frac{d^2y}{dx^2} = \frac{f''(t)}{1} = f''(x). Every graph is a parametric curve with the parameter equal to x.