AP Calculus BC Topic 9.2: Second Derivatives of Parametric Equations

Topic 9.2 key terms

parametric equations: A pair of functions $x(t)$ and $y(t)$ of one parameter $t$ (often time). Each value of $t$ gives one point $(x(t), y(t))$ of the path.
slope of a parametric curve: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ , provided $dx/dt \neq 0$ : the slope of the line tangent to the curve at the point given by $t$ .
tangent line at a time t: The line through $(x(t_0), y(t_0))$ with slope $\left.\frac{dy}{dx}\right|_{t=t_0}$ ; point and slope both come from the same $t_0$ .
horizontal tangent: Where $dy/dt = 0$ and $dx/dt \neq 0$ : the path is momentarily neither rising nor falling (the object need not be stopped).
vertical tangent: Where $dx/dt = 0$ and $dy/dt \neq 0$ : $dy/dx$ is undefined and the tangent line is vertical. If both are 0, the quotient alone can't decide.
second derivative of a parametric curve: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$ : how fast the slope changes per unit of $x$ . It is not $y''(t)/x''(t)$ .
concavity of a parametric curve: Concave up where $d^2y/dx^2 > 0$ , concave down where $d^2y/dx^2 < 0$ , read along increasing $x$ (so the sign of $dx/dt$ matters).
ricochet (skip) rule: An engineers' rule of thumb: a sphere skips off water if it meets it at less than about $18°/\sqrt{\text{specific gravity}}$ , about 6.7° (a slope of about 0.12) for cast iron.

Transcript

1,227 words · 4 parts

How fast the slope turns

Back at the ball, let’s read its slope once every second. It goes one point one seven, zero point seven, zero point two three, and keeps falling, all the way down to negative one point one seven. Gravity takes nine point eight off dy dt every second while dx dt stays twenty-one, so the slope drops by nine point eight over twenty-one, about zero point four seven, every single second.

The second derivative asks how fast the slope changes for every meter you move along x, instead of every second. The ball covers twenty-one meters of x every second, so per meter the slope changes by negative zero point four seven divided by twenty-one. That’s about negative zero point zero two two, and that number is d two y d x squared.

You might try just differentiating the top and bottom again, y double prime over x double prime, so let’s test that idea on this flight. This ball slides sideways at a steady twenty-one meters per second, so x double prime is zero, and you’d get negative nine point eight over zero, which is nonsense for a path that’s so obviously curving.

Now suppose we fire the ball at double the speed, still aimed along the laser. Per meter of x, how fast do you think its slope will change, compared with now?

To get d two y d x squared, take the slope dy dx, differentiate it with respect to t, and then divide by dx dt, which turns per second into per unit of x. For the ball, call the slope m of t. It’s twenty-four point five minus nine point eight t, over twenty-one, which simplifies to seven sixths minus seven fifteenths t.

Its t-derivative is negative seven fifteenths, the same negative zero point four seven per second we read off the tags, and dividing by twenty-one gives negative one forty-fifth, the negative zero point zero two two per meter from before. On the AP exam you’d finish with a sentence like this one. Because d two y d x squared equals negative one forty-fifth, which is negative for every t, the path is concave down everywhere.

Let’s do one where dx dt isn’t constant, x equals t squared and y equals t cubed, for positive t. The slope is three t squared over two t, which is three t over two.

Differentiate that in t and you get three halves, then divide by dx dt, which is two t, to get three over four t. At t equals one that’s three quarters, which is positive, so the curve is concave up at the point one, one. The shortcut, y double prime over x double prime, is six t over two, which gives three, four times too big.

Try one yourself, with x equals t squared and y equals four t. What’s d two y d x squared at t equals one?

Her shot

Back on the Atlantic, the enemy fires back at us. Her ball rises as it flies from right to left, so read from left to right, its path slopes down at first. Its slope goes negative zero point six seven, then negative zero point five three, then negative zero point four. Every second her slope goes up by two fifteenths, about zero point one three, until it splashes down, well short of us, at about positive zero point six eight.

Her slope goes up every second, so what do you think that tells you about the concavity of her path? Now test your answer with your own hands. Drag the marker along the water from left to right, and watch what her slope does as x increases.

Side by side, the two shots are mirror images, both arching down. Our slope falls by two fifteenths every second while we move right, and hers rises by two fifteenths while she moves left. Dividing by dx dt turns per second into per foot, and it also keeps track of which way x is going.

Will it skip?

Now let’s do that cannonball problem the way the AP exam would ask it. The ball leaves a deck sixteen feet above the sea, with these two equations. We’ll use the engineers’ cutoff from the start, so it skips if the size of its slope at the water is less than zero point one two, a line tilted about seven degrees. Does this one skip? First we need to know when it reaches the water, which is when y of t is zero.

Dividing everything by negative eight gives two t squared minus three t minus two, which factors nicely. That gives t equals two seconds, and we throw out negative one half because that’s before the gun even fired. At that moment dx dt is four hundred feet per second and dy dt is twenty-four minus sixty-four, which is negative forty, so the slope is negative forty over four hundred, negative zero point one.

Then write the sentence a grader wants to see. At t equals two, dy dx is negative zero point one, and its size is less than zero point one two, so the ball meets the water shallow enough to skip. The inverse tangent of zero point one is about five point seven degrees, so that’s the angle it hits at.

Now you try it with less powder and the barrel tipped up a touch, so x of t is three hundred t and y of t stays the same. What’s the slope dy dx when this ball reaches the water, to three decimal places? Let’s see if the sea agrees. The four hundred shot skips right off the water, and the three hundred shot comes in just a little too steep and digs in.

Now you’re the gun captain. She has moved closer, to about fifteen hundred feet. Set the elevation and the powder, tilt the line to predict the slope at the water, and then fire to skip a ball into her side. That’s ricochet fire, and now you can predict the angle it comes in at before the gun even fires.

On the exam

On calculator-active parts of the exam, they often hand you the two rates directly. Here dx dt is the square root of t plus one, and dy dt is the natural log of one plus t squared. What’s the slope of its path at t equals three? Use the calculator and give it to three decimal places.

A flying cannonball is a pair of parametric equations, two numbers driven by one clock, and the direction it’s heading at any instant is its climb rate divided by its sideways rate. The tangent goes flat when the climb rate is zero and the sideways rate isn’t, and it stands straight up when it’s the other way round. To find how fast that slope changes for every foot of x, differentiate it in t and divide by dx dt once more.

At the start I promised you’d be able to tell, before a cannonball ever touches the water, the angle it’ll hit at and whether it skips, and you just did it, on paper and on the gun. The next time you skip a stone across a lake, throw it low and fast so it meets the water at a shallow slope, and give it a spin to help keep its flat face steady.

Topic 9.2: Second Derivatives of Parametric Equations