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### Angolo ottimale per un proiettile parte 4 Ricerca del ottimale angolo e la distanza con un po' di matematica

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Now that we have distance explicitly as a function of
the angle that we're shooting the object at, we can use a
little bit of calculus to figure out the optimal angle,
the angle that's going optimize our distance.
And since we only care about angles from 0 degrees to
really 90 degrees, let's constrain ourselves.
So we're going to optimize things for
angles between 0 degrees.
So theta is going to be greater than or equal to 0 and
less than or equal to 90.
So let's see how we can do it.
And just to get an idea of what we're even conceptually
doing with the calculus, remember when you take a
derivative, you are finding the slope of a line, an
instantaneous slope of a line.
And if you were to graph this-- and I encourage you to
graph it on your own, maybe with a graphing calculator--
it will look something like this over the interval.
It will look like this where that is the distance as a
function of theta axis and then this would
be our theta axis.
And we care about angles between 0 and 90 degrees.
So if you were to graph this thing, so this is 0 degrees,
this is maybe 90 degrees right here.
The graph of this function will look like this.
It'll look something like this.
It will look something like that.
And what we want to do is find the angle, there's some angle
here that gives us the optimal distance.
So this is, right here, this is the optimal distance.
And what we want to do is find that out.
And when you look at the graph, and you could do it on
a graphing calculator if you like, what happens to the
instantaneous slope at that optimal distance?
Well it's flat.
The slope there is 0.
So what we need to do is take the derivative of this
function and then just figure out at what angle is the
derivative or the instantaneous slope of this
function equal to 0?
And then we're done We will know this mystery angle, this
optimal angle, to shoot the object at.
So let's take the derivative.
So the derivative, we'll just use our derivative rules here.
The derivative of-- I will call it d prime I guess, or we
could say the derivative of the distance with respect to
theta is equal to-- we're assuming that s and g are
constants, so we don't have to worry about them right now.
We could just put them out front since we're assuming
they're constants.
And then we can do the product rule to take the derivative of
this part with respect to theta.
In the product rule, we take the derivative of the first
function times the second function.
So the derivative of cosine of theta is
negative sine of theta.
And we're going to multiply that
times the second function.
So that's times the sine of theta.
And to that we're going to add the first function, which is
cosine of theta times the derivative
of the second function.
The derivative of sine theta is cosine of theta.
I know it's a little bit confusing.
All we did is we took the derivative of the first one
times the second one.
And then we took the derivative of the second one
times the first one.
Let me make it even more explicitly clear.
We took the derivative of this guy here, so this is the
derivative with respect to theta.
And we took the derivative of this guy over here with
respect to theta.
We took the derivative of cosine there and
multiplied it by sine.
Took the derivative of sine here and
multiplied it by cosine.
Just the product rule.
Now what does this give us?
We can simplify this a good bit.
So we could write the derivative d prime is equal
to-- we could keep this constant out there-- 2s
squared over g-- times-- now negative sine of theta times
sine of theta, that's just negative
sine squared of theta.
And then, cosine theta times cosine theta, that's just plus
cosine squared of theta.
Now, what we just said is we want to figure out the point,
the angle at which this derivative or the
instantaneous slope is 0.
So let's set this thing equal to 0.
So we just have to solve for theta now.
Now the first thing I do to solve for theta is just divide
both sides by 2s squared over g.
If you divide the left-hand side by that, it cancels out
with 2s squared over g.
And if you divide 0 by that, assuming this isn't 0, which
it shouldn't be, then you'll still get 0.
So this equation simplifies to-- I'll write it in blue--
negative sine squared of theta plus cosine squared of theta
is equal to 0.
Now, if we add sine squared of theta of both sides of this
equation, let's add sine squared of
theta to both sides.
We are left with-- these cancel out.
Cosine squared of theta is equal to
sine squared of theta.
Now, both of these are going to be positive over the
interval, so we're going to just take the positive square
root of both of them, or the principal root of both sides
of this equation.
So let's do that.
So you take the principal roots of both
sides of this equation.
You could do it that way.
Actually, a more interesting way than doing it that way, is
to divide both sides of this equation by cosine squared of
theta assuming that it's not equal to 0 over this interval.
So cosine squared of theta.
You could also do it using the positive square root, the
principal root, either one will work.
But this is interesting because the left-hand side
cancels out to 1, and the 1 will be equal to-- what's sine
squared over cosine squared of theta?
Well that's the same thing as sine of theta over cosine of
theta squared.
You have a square divided by another square.
That's the same thing as the numerator divided by the
denominator.
That whole thing squared.
And what's sine of theta divided by cosine of theta?
Well that's just the tangent of theta.
So we have 1 is equal to the tangent squared of theta.
Or, we could take the positive square root of both sides of
this equation.
Tangent is positive over the interval from 0 to 90 degrees,
so that's cool to do.
So if you take the positive square root of both sides, you
get the positive square root of 1 is 1.
1 is equal to tangent of theta.
And then you take the inverse tan of both sides or the arc
tan of both sides and you get the arc tan of
1 is equal to theta.
And this is just a very fancy way of saying theta's the
angle that if you were take its tangent, you get 1.
And you could use a calculator to solve that or you might
just know that by memory.
This theta, the arc tangent of 1 is 45 degrees.
Or if you are dealing in radians, it
Either one of those is going to work.
So our optimal angle when we shoot this thing is going to
be at 45 degrees.
Now, what is that optimal distance going to be when we
shoot it off at 45 degrees?
Well, we can just go back to our original formula.
We just go back to our original
formula that we derived.
If we're shooting it off at 45 degrees, what is the sine of
45 degrees?
The sine of 45 degrees is equal to the square
root of 2 over 2.
You could use a calculator for that or maybe you know it from
the unit circle.
The cosine of 45 degrees is also square root of 2 over 2.
And if you'd actually just taken the principal roots at
this stage of the equation, you'd have gotten that the
cosine of theta has to equal sine of theta over this
interval, and that only happens at 45 degrees.
But given this.
We can put this back into the original expression right up
here, our original function.
So the optimal distance that we are going to travel, so
distance as a function-- the distance we travel at 45
degrees is going to be equal to 2s squared over g times
cosine of theta, which is square root of 2 over 2.
Cosine of 45 is square root of 2 over 2 time sine of theta,
which is square root of 2 over 2.
Well what's the square root of 2 times the square root of 2?
Well that's just 2.
Let me simplify this.
So the square root of 2 times the square root
of 2, that is 2.
This 2 cancels out with that 2.
And then, this 2 cancels out with this 2.
So then the optimal distance you travel at 45 degrees, all
we're left with is the s squared over g.
Assuming no air resistance, kind of an ideal circumstance.
No matter what planet you're on, how fast you do it, the
best angle is always 45 degrees assuming no air
resistance.
And if you do it on that best angle, you're going to travel
s squared over g.
Going back to the original problem, if s is
10 meters per second.
Let's say s was 10 meters per second.
And let's say we're dealing with a world where let's say,
gravity is equal to 10 meters per second squared, then
according to what we've derived, your optimal distance
is going to be s squared-- so it's going to be 100-- divided
by gravity.
It's going to be 10.
And if you square meters per second, you're going to get
meters squared per second squared divided by the
acceleration of gravity, meters per second squared.
Second squared's cancel out.
You have a meter squared divided by meters.
Your optimal distance would be 10 meters.
Pretty neat.