Topic 9.5 key terms
- position vector
- r(t) = ⟨x(t), y(t)⟩: the arrow from the origin to the moving object at time t.
- derivative of a vector-valued function
- r′(t) = ⟨x′(t), y′(t)⟩: differentiate one component at a time; every derivative rule (chain rule included) works on each component.
- velocity vector
- v(t) = r′(t). It sits on the object and points the way it is moving; its components are how fast x and y are changing.
- acceleration vector
- a(t) = v′(t) = r″(t): how the velocity is changing, in size or in direction. Moving in a circle at constant speed still has acceleration, toward the center.
- speed
- The length of the velocity: |v(t)| = √(x′(t)² + y′(t)²). A number, never negative; speed is not a vector.
- initial condition
- One known value (a position or a velocity at one time) that fixes the constants of integration, one per component.
- particular solution
- The one position function with the given velocity that passes through the known point: r(t) = r(t₀) + ∫ from t₀ to t of v(u) du.
- displacement
- ∫ from a to b of v(t) dt = r(b) − r(a): the net change in position over [a, b], a vector (how far and which way).
- total distance traveled
- ∫ from a to b of |v(t)| dt = ∫ from a to b of √(x′(t)² + y′(t)²) dt (a ≤ b): the integral of the speed, the length of the path actually traveled.
Transcript
729 words · 1 parts
Running it backwards
Now let’s run calculus backwards, somewhere with a lot more fire. This is Stromboli, a volcano off the north coast of Sicily that has been almost continuously active for more than two thousand years, throwing glowing rock above its vents a few times every hour. Each glowing rock flies its own path at its own speed, but once it leaves the vent, nothing pushes it sideways and gravity pulls it straight down.
That gives every rock the same acceleration, zero, negative nine point eight one meters per second squared, at every point of its flight, even at the top. If all we know is that acceleration and how this one rock left the vent, can we rebuild its whole flight? Researchers filmed Stromboli with thermal cameras, and in its rock-throwing explosions the rocks crossed the crater rim at about thirty-four meters per second on average. Let’s give ours exactly that, sixteen across and thirty up, because sixteen squared plus thirty squared is thirty-four squared.
Going from acceleration back to velocity means integrating, one component at a time. Zero integrates to a constant, and negative nine point eight one integrates to negative nine point eight one t plus another constant. The launch tells us both constants, because at t equals zero the velocity is just C one, C two, which has to be sixteen, thirty.
The velocity is sixteen, thirty minus nine point eight one t, so the sideways part never changes, and the upward part loses nine point eight one every second. Integrate again for the position, which gives you sixteen t, thirty t minus four point nine oh five t squared, plus two new constants.
We put the vent at the origin, so both new position constants are zero, and the whole flight is sixteen t, thirty t minus four point nine oh five t squared. You may recognize that second part from physics, v zero t minus one half g t squared. We just derived it ourselves, by integrating gravity twice and using the launch velocity and the launch position.
Let’s see whether the rock agrees. The dashed line is our formula’s prediction, with a tick every half second. It flies right along the dashes, crossing every tick on time, and lands about ninety-eight meters from the vent, about six point one seconds after launch.
Now you get to choose the constants yourself, so pick a launch velocity that lands a rock in the ring sixty meters away, and let the dashed path show you where it’ll land before you fire. Let’s go back to our original rock, the one launched at sixteen, thirty, so take hold of it, drag it along its flight, and stop it at the moment it’s moving the slowest.
The volcano worked because we knew where the rock started, and every problem like this needs one known point to pin the path down. Let’s try one where the known time isn’t zero. The velocity is two t, negative two, and at t equals one the particle is at three, four, so where is it at t equals three?
I want position, so I integrate each velocity component. Two t gives t squared plus C one, and negative two gives negative two t plus C two. At t equals one that’s one plus C one, negative two plus C two, which has to equal three, four, so C one is two and C two is six.
That makes r of t equal to t squared plus two, six minus two t, and at t equals three the particle is at eleven, zero. There’s a faster way that skips the constants, because the position at any time is the known position plus the integral of the velocity from the known time, where u is just the time inside the integral.
Here that’s three, four plus the integrals from one to three, eight and negative four, which lands at eleven, zero again. Two wrong answers show up all the time. Stop at eight, negative four, and you’ve found the displacement but skipped the starting point. Get nine, negative six, and you plugged in three without using the known point to find the constants.
Now you try one. The velocity is three t squared, two t minus four, and at t equals one the particle is at two, five. Where is it at t equals three?