Topic 3.4 key terms
- derivative of an inverse function
- If f is differentiable and one-to-one, f(a) = b and f′(a) ≠ 0, then (f⁻¹)′(b) = 1/f′(a): find the input that gave the output, then take the reciprocal of the slope there.
- reciprocal slopes at mirrored points
- Reflecting a graph across y = x swaps each point (a, b) to (b, a) and swaps the rise and run of every tangent triangle, so the slopes there multiply to 1.
- provided the derivative exists
- Where f has a horizontal tangent (f′(a) = 0), its inverse has a vertical tangent at b = f(a) and no derivative there, as with x³ and ∛x at 0, or arcsin at ±1.
- derivative of arcsin
- d/dx arcsin x = 1/√(1 − x²) for −1 < x < 1; with the chain rule, d/dx arcsin(u) = u′/√(1 − u²).
- derivative of arccos
- d/dx arccos x = −1/√(1 − x²) for −1 < x < 1; with the chain rule, d/dx arccos(u) = −u′/√(1 − u²).
- derivative of arctan
- d/dx arctan x = 1/(1 + x²) for every real x; with the chain rule, d/dx arctan(u) = u′/(1 + u²).
- angle of a tracking camera
- A camera d away from a vertical climb at height h has angle θ = arctan(h/d), so θ′ = d·h′/(d² + h²) radians per second.
Transcript
1,427 words · 5 parts
Arcsine and arccosine
Now let’s use this on the inverse trig functions, starting with arcsine. y equals arcsine of x means sine of y equals x, with y between negative pi over two and pi over two. You’ll also see arcsine written as sine with a little negative one, and that symbol asks which angle has this sine, which is a different thing from one over sine.
Differentiating both sides of sine y equals x, the way you did with implicit differentiation, gives cosine of y times dy dx equals one, so dy dx is one over cosine of y. That’s the mirror’s rule again, because cosine of y is sine’s slope at y, the input that sine sends to x.
To write that in terms of x, use sine squared plus cosine squared equals one. Cosine of y is the square root of one minus x squared. It’s the positive root, because on this branch y stays on the right half of the unit circle, where cosine is never negative. That’s the derivative of arcsine, one over the square root of one minus x squared, and I love that there’s no trig left in it at all.
At the ends, sine is flat, so its mirror image, arcsine, stands straight up at one and negative one, and the derivative blows up there, just like the cube root’s did. Arccosine works the same way. Cosine of y equals x, with y from zero to pi, gives negative sine of y times y prime equals one.
That makes y prime negative one over sine of y, and on this branch sine of y is the square root of one minus x squared, so the derivative is negative one over the root of one minus x squared. That’s arcsine’s derivative with a minus sign, which makes sense, because arccosine of x is pi over two minus arcsine of x.
Call the inside u. The chain rule puts u prime on top and squares the whole u under the root, and arccosine gets the same thing with its minus sign. For arcsine of x cubed, that’s three x squared over the root of one minus x to the sixth.
The mistake I see most is leaving one minus x squared under the root, as if the inside were still just x. Now you try one. Which of these is the derivative of arccosine of x squared?
The whole sky in a circle
Let’s take arcsine somewhere prettier than a whiteboard. We’re lying on the sand at the bottom of a clear lagoon, four meters down, about to look straight up. The whole sky, from horizon to horizon, fits inside this one bright circle, and outside it the surface turns into a mirror. Divers call it Snell’s window.
It happens because light bends when it enters water. A ray coming down at angle alpha from straight up ends up at a smaller angle beta, and sine of alpha equals one point three three times sine of beta.
Turn that around, and when you look up at angle beta, you see the sky at alpha equals arcsine of one point three three sine beta. Sky rays that start out ten degrees apart crowd closer and closer together under the water as they near the edge.
Let’s make a map of the sky. Tilt your view until you’re looking at sky thirty degrees from straight up. Now find the sky eighty degrees from straight up, just ten degrees above the horizon.
Look at your map. The first thirty degrees of sky took twenty-two degrees of tilt, the next fifty took only about twenty-six, and the last ten above the horizon fit into less than one degree. The window ends at about forty-nine degrees, where one point three three sine beta reaches one. Past that, no angle has a sine that big, so arcsine has no value there, no sky light reaches you, and you see a mirror.
The edge of the window is the edge of arcsine’s domain, right over your head. To find how fast the sky slides by as you tilt, the chain rule puts one point three three cosine beta on top and squares the whole inside under the root. Use the calculator to find that rate at forty-five degrees, to the nearest hundredth.
Near the rim the slope runs away. At forty-eight degrees the sky moves almost six times as far as your gaze, and that rate keeps growing without any limit as you reach the edge.
Arctangent
The last one is the one our camera needs. y equals arctangent of x means tangent of y equals x, with y strictly between negative pi over two and pi over two. Differentiating both sides gives secant squared of y times y prime equals one, so y prime is one over secant squared of y.
Secant squared is one plus tangent squared, and tangent of y is x, so the bottom is one plus x squared. You can also read that off this triangle with legs one and x. The derivative of arctangent is one over one plus x squared, and it works for every real x, with no endpoints at all.
Arctangent is steepest at zero, where its slope is one, and it flattens out toward negative pi over two on the left and pi over two on the right. With something inside, the derivative of arctangent of u is u prime over one plus u squared.
Back to the camera. It’s d kilometers from the pad and the rocket is h kilometers up, so tangent of the angle is h over d, and theta is arctangent of h over d. Only h changes, so the inside’s derivative is h prime over d. Multiplying top and bottom by d squared, one d cancels, and that leaves d times h prime over d squared plus h squared, in radians per second.
Keep it in the frame
Let’s try it at the eight kilometer ring, where the camera’s triangle is six, eight, ten and the rocket climbs about zero point two seven seven kilometers per second. On top, six times zero point two seven seven is about one point six six, and on the bottom, thirty-six plus sixty-four is exactly a hundred. The camera turns about zero point zero one six six radians per second, just under a degree every second.
Now it’s your camera. Forty seconds after liftoff the rocket is three point six four kilometers up, climbing zero point one seven three kilometers per second, so work out the rate, set the motor, and press run.
Now keep that same motor rate and jump to ninety seconds, when the rocket is three times faster. The camera turns far too fast, and the rocket sinks out of the frame.
That rate only works at forty seconds. At ninety seconds the rocket is twenty kilometers up, climbing zero point five three six kilometers per second, so work out the new rate and set the motor again.
Let’s draw the camera’s turning rate for the whole climb and pin your two settings on it. Your forty-second setting sits right at the top of this curve, the fastest this camera ever turns, and your ninety-second setting is about a third of it. Now we can explain the surprise. The formula is two rates multiplied together, d over d squared plus h squared, the turn for each kilometer of climb, times h prime, the kilometers per second.
From forty seconds to ninety, the climb rate triples, but the turn for each kilometer falls to about a ninth of what it was, because h squared on the bottom keeps growing. Multiplying those two changes leaves about a third of the turning rate. Real rockets also travel downrange, out over the ocean, so a real tracking camera has to follow that sideways motion too, and our straight-up model isolates the part we just worked out.
What the camera knew
Every derivative in this lesson came from one move, writing the inverse in terms of the original function and differentiating both sides. For any inverse, find the input first, then flip the slope there. For arcsine that gave one over the root of one minus x squared, the same with a minus sign for arccosine, and for arctangent, one over one plus x squared.
With arctangent you can now tell this camera exactly how fast to turn at any second of the flight, and why it slows down while the rocket speeds up. Next time you watch a launch, look for the tracking shot, because now you know the calculus that keeps the rocket in the frame.