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## Optimal angle for a projectile

# Optimal angle for a projectile part 4: Finding the optimal angle and distance with a bit of calculus

## Video transcript

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 is pi over 4 radians. 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.