# Math Help - find function and solve max problem

1. ## find function and solve max problem

Suppose this fox wants to try and kill a rabbit. He has a catapult which is capable of launching a rock at 70 m/s. The fox found a completely flat spot on cartoon land (which he labelled as X). He wants to place the catapult so that it is due west of X. Further, he wants to try get as far away from X as he can to be safe. The only force acting on the rock is gravity; on cartoon land, it has a constant downward component of 8.75 m/s^2, and a constant westerly component of 3.5 m/s^2. There is no southerly component to take into account.

1.) What's the angle from the horizontal that the fox should aim the catapult in order to maximize the distance that the rock will travel. Give the max. distance. What is the plot of the path that the rock will go on?

2.) What's the angle the fox should avoid to prevent getting hit by the rock. Give the plot of this path that the fox does NOT want the rabbit to take.

2. Originally Posted by flash101
Suppose this fox wants to try and kill a rabbit. He has a catapult which is capable of launching a rock at 70 m/s. The fox found a completely flat spot on cartoon land (which he labelled as X). He wants to place the catapult so that it is due west of X. Further, he wants to try get as far away from X as he can to be safe. The only force acting on the rock is gravity; on cartoon land, it has a constant downward component of 8.75 m/s^2, and a constant westerly component of 3.5 m/s^2. There is no southerly component to take into account.

1.) What's the angle from the horizontal that the fox should aim the catapult in order to maximize the distance that the rock will travel. Give the max. distance. What is the plot of the path that the rock will go on?
Step 1. Draw a sketch. I can't do this, but I can explain what I have in front of me. I've got the catapult on the ground facing west so that the acceleration due to gravity will help me fire the rock the furthest. The catapult is situated on the ground and the ground is level to the point where the rock hits.

Step 2. Set up a coordinate system. There is a slightly bizzare feature here: the simplest coordinate system is not the most natural one. I will go with natural and choose a +x direction to the West and a +y direction straight up. I am going to choose an origin located at the point where the rock is fired. (Technical point here: I am assuming the rock is fired at the origin so technically the base of the catapult is underground. This is a standard assumption to make the problem simpler.)

Step 3. List all the given information. I usually use some kind of table, like the following:
t0 = 0 s................................t = ?
x0 = 0 m....x = ?..................y0 = 0 m.......y = 0 m
v0x = ?......vx = ?.................v0y = ?.........vy =?
.....ax = 3.5 m/s^2...................ay = -8.75 m/s^2

Let me explain this. We are firing the rock from the origin (the "0" variables) so x0 = 0 m and y0 = 0 m, and it is landing on the ground (which is at y = 0 m). We are firing at time 0 t0 = 0 s. We don't know when the rock lands so t is unknown. We don't know what angle we are launching at, but we can say that
v0x = (70)cos(@) m/s
v0y = (70)sin(@) m/s
where @ is the angle of inclination of the launch.

Step 4. Set up the equations.
We have 8 equations at our disposal:
x = x0 + v0x*t + (1/2)ax*t^2
x = x0 + (1/2)(v0x + vx)t
vx = v0x + ax*t
vx^2 = v0x^2 + 2ax*(x - x0)

y = y0 + v0y*t + (1/2)ay*t^2
y= y0 + (1/2)(v0y + vy)t
vy = v0y + ay*t
vy^2 = v0y^2 + 2ay*(y - y0)

We wish to maximize the range, so a likely equation to use is the first x equation:
x = x0 + v0x*t + (1/2)ax*t^2
x = 70t*cos(@) + 1.75t^2

We need to maximize x with respect to @, but we need an equation for t first. So let's look for a y equation.

y = y0 + v0y*t + (1/2)ay*t^2
0 = 0 + 70t*sin(@) - 4.375t^2

or
4.375t = 70*sin(@)
t = (70/4.375)*sin(@)

Putting this into our x equation:
x = 70t*cos(@) + 1.75t^2

x = 70[(70/4.375)*sin(@)]*cos(@) + 1.75[(70/4.375)*sin(@)]^2

x = 1120*sin(@)*cos(@) + 448*sin^2(@)

To find the max x we need to vary this with respect to @. So take the first derivative of x with respect to @ and set it to 0:
0 = 1120*cos^2(@) - 1120*sin^2(@) + 2*448*sin(@)*cos(@)

Analytically speaking, this is complicated to solve. If you are allowed, I would recommend you use a calculator to solve this. Otherwise I would start by noting that @ is in the first quadrant, so we may put:
cos(@) = +sqrt(1 - sin^2(@))

So the equation is:
0 = 1120*(1 - sin^2(@)) - 1120*sin^2(@) + 896*sin(@)*sqrt(1 - sin^2(@))

2240*sin^2(@) - 1120 = 896*sin(@)*sqrt(1 - sin^2(@)) <-- Square both sides.

(2240*sin^2(@) - 1120)^2 = 802816*sin^2(@)*(1 - sin^2(@))

5017600*sin^4(@) - 5017600*sin^2(@) + 1254400 = 802816*sin^2(@) - 802816*sin^4(@)

5820416*sin^4(@) - 5820416*sin^2(@) + 1254400 = 0

To make this look a little simpler, set z = sin^2(@):
5820416*z^2 - 5820416*z + 1254400 = 0

z^2 - z + 0.215517 = 0

z = (1 (+/-) sqrt((-1)^2 - 4*1*0.215517))/(2*1)

z = 0.685695 or z = 0.314305

So
sin(@) = (+/-)sqrt(z)
sin(@) = 0.82867 or sin(@) = 0.560629 <-- I'm only choosing the + solutions since @ is in QI.

Thus @ = 55.9007 degrees or @ = 34.0993 degrees.

As to what the plot looks like, look at the x equation again:
x = 70t*cos(@) + 1.75t^2

@ is just a constant, so this will be a parabola on a t vs. x plot.

-Dan

3. Originally Posted by flash101
2.) What's the angle the fox should avoid to prevent getting hit by the rock. Give the plot of this path that the fox does NOT want the rabbit to take.
I don't have the time to do this one, but the method is the same as the last one. The difference is that the final point for the rock is now x = 0 m and y = 0 m. This makes the solution of the equation for @ a bit simpler.

-Dan

4. I'm being silly. There is an obvious way to approach the second problem, but I discarded it due to the coordinate system.

Shortcut method: The path back to the origin is going to have to be a straight line (can you see why?) So the rock will have to be projected in the opposite direction to the overall acceleration.

So add the two acceleration components. This will give you the direction of the acceleration with respect to the coordinate system I used in the first problem. (i.e. find the angle between the net acceleration and the Westerly component of the acceleration.) You want to project the rock in the exact opposite direction.

-Dan

5. I'm very impressed, Dan. No wonder they call you the physics maestro. I'll take a look through all of your work. My prof. did say this was very complicated, which explains your extensive work. Many thanks.

6. Hi Dan,

In order to try and understand your work I should first try understand the 'sketch' you tried describing.

You have the catapult facing WEST, which you said will help fire the rock the furthest.

So, something like:

Code:

<***                         X
So x is where the rabbit will be; the catapult is the *** with the < to indicate it is facing west. How will that ever reach X.

Next, when you talk about your coordinate system, I am assuming you're just using the standard x-y system, not incorporating z anywhere in there.

The part where you have t_0 = 0s, at time t = 0, the rock has gone no where, and then you have t = ?. That's how I interpreate it, although the other info. with t = ? is confusing me. I understand a_x = 3.5 m/s^2 is the constant westerly component; and similary a_y = -8.75 m/s^2 is the constant downward component (I assume you have negative because its DOWNWARD? but in the problem it gives it as positive). You mention t is unknown because we don't know when it will hit the ground...I understand this part.

How'd you come up with v_0x = 70cos(@) m/s. v_0y = 70sin(@) m/s?

The rest obviously will follow from understanding the above; some of it looks quite complicated.

Just a brief background of what I've learned in multi thus far that might help:

Dot/cross products and since ST ||vector PQ x vector PR||/||vector PR|| gives the distance between them (where x is the cross product).

Projection/components (Proj_b(a)/comp_b(a)) etc.

Arclength etc etc.

I guess the big one relating to this is finding different vectors and then integrating/finding derivatives of these vector-valued func.'s to solve them, which I think is what you sort of have.

Any way, I'll continue sifting through what you've shown. Thanks- also if you could clarify the above for me. It just seems weird that the catapult will be facing west when it's west already.

7. Originally Posted by flash101
You have the catapult facing WEST, which you said will help fire the rock the furthest.

So, something like:

Code:

<***                         X
So x is where the rabbit will be; the catapult is the *** with the < to indicate it is facing west. How will that ever reach X.
The acceleration has a component in the Westerly direction. So any object, say, fired straight up into the air will travel West. Hence we want the catapult facing West toward a target to the West. Now presumably the target could be to the East, which I didn't take into consideration, but the question is asking for the maximum possible range, which can thus only occur if the rock is fired to the West.

Originally Posted by flash101
Next, when you talk about your coordinate system, I am assuming you're just using the standard x-y system, not incorporating z anywhere in there.
Yes. The problem doesn't mention anything about a third direction, so I didn't worry about it.

Originally Posted by flash101
The part where you have t_0 = 0s, at time t = 0, the rock has gone no where, and then you have t = ?. That's how I interpreate it, although the other info. with t = ? is confusing me.
The "t" variable in this case is representing the point in time when the rock hits the ground. (Sorry for the confusion. Typically t0 is defined to be 0 s, and none of the equations of motion explicitly use t0, but there are occasions where it is more convenient to use a t0 that isn't 0 s. That's why I include it in the table.)

Originally Posted by flash101
I understand a_x = 3.5 m/s^2 is the constant westerly component; and similary a_y = -8.75 m/s^2 is the constant downward component (I assume you have negative because its DOWNWARD? but in the problem it gives it as positive).
This is a typical confusion for many students. In 1 dimension the direction vectors take is denoted by whether the "component" of the vector in that direction is positive or negative. So given a set of components for the acceleration, which components should be positive and which should be negative?

The answer is that this depends entirely on the coordinate system you choose. (Which is why I ALWAYS list the coordinate system I am using.) I chose +x to be to the West, so ax is positive since it is in this direction. I chose +y to be upward, so ay is negative since it is in the downward (-) direction.

Originally Posted by flash101
How'd you come up with v_0x = 70cos(@) m/s. v_0y = 70sin(@) m/s?
These are the vector components of the initial velocity. I assumed the initial velocity was 70 m/s (given in the problem) at an angle of @ above the +x axis.

Originally Posted by flash101
some of it looks quite complicated.
The Mathematics involved with solving the problem are a bit on the complicated side. However, you should be able to easily understand the Physics, which is how I set up those (8) equations. As far as the Math is concerned some problems (like this one) you have to see solved once before you can solve one on your own.

Originally Posted by flash101
Just a brief background of what I've learned in multi thus far that might help:

Dot/cross products and since ST ||vector PQ x vector PR||/||vector PR|| gives the distance between them (where x is the cross product).

Projection/components (Proj_b(a)/comp_b(a)) etc.

Arclength etc etc.

I guess the big one relating to this is finding different vectors and then integrating/finding derivatives of these vector-valued func.'s to solve them, which I think is what you sort of have.

Any way, I'll continue sifting through what you've shown. Thanks- also if you could clarify the above for me. It just seems weird that the catapult will be facing west when it's west already.
I have approached the solution to this problem in terms of how I would solve the Physics problem. From the sounds of it you are in a Mathematics class so you might have a different set of notations for what I worked out. If so, I suggest that you check with your Professor about how (s)he would word my solution so you can understand it better.

-Dan

8. Originally Posted by topsquark
Step 1. Draw a sketch. I can't do this, but I can explain what I have in front of me. I've got the catapult on the ground facing west so that the acceleration due to gravity will help me fire the rock the furthest. The catapult is situated on the ground and the ground is level to the point where the rock hits.

Step 2. Set up a coordinate system. There is a slightly bizzare feature here: the simplest coordinate system is not the most natural one. I will go with natural and choose a +x direction to the West and a +y direction straight up. I am going to choose an origin located at the point where the rock is fired. (Technical point here: I am assuming the rock is fired at the origin so technically the base of the catapult is underground. This is a standard assumption to make the problem simpler.)

Step 3. List all the given information. I usually use some kind of table, like the following:
t0 = 0 s................................t = ?
x0 = 0 m....x = ?..................y0 = 0 m.......y = 0 m
v0x = ?......vx = ?.................v0y = ?.........vy =?
.....ax = 3.5 m/s^2...................ay = -8.75 m/s^2

Let me explain this. We are firing the rock from the origin (the "0" variables) so x0 = 0 m and y0 = 0 m, and it is landing on the ground (which is at y = 0 m). We are firing at time 0 t0 = 0 s. We don't know when the rock lands so t is unknown. We don't know what angle we are launching at, but we can say that
v0x = (70)cos(@) m/s
v0y = (70)sin(@) m/s
where @ is the angle of inclination of the launch.

Step 4. Set up the equations.
We have 8 equations at our disposal:
x = x0 + v0x*t + (1/2)ax*t^2
x = x0 + (1/2)(v0x + vx)t
vx = v0x + ax*t
vx^2 = v0x^2 + 2ax*(x - x0)

y = y0 + v0y*t + (1/2)ay*t^2
y= y0 + (1/2)(v0y + vy)t
vy = v0y + ay*t
vy^2 = v0y^2 + 2ay*(y - y0)

We wish to maximize the range, so a likely equation to use is the first x equation:
x = x0 + v0x*t + (1/2)ax*t^2
x = 70t*cos(@) + 1.75t^2

We need to maximize x with respect to @, but we need an equation for t first. So let's look for a y equation.

y = y0 + v0y*t + (1/2)ay*t^2
0 = 0 + 70t*sin(@) - 4.375t^2

or
4.375t = 70*sin(@)
t = (70/4.375)*sin(@)

Putting this into our x equation:
x = 70t*cos(@) + 1.75t^2

x = 70[(70/4.375)*sin(@)]*cos(@) + 1.75[(70/4.375)*sin(@)]^2

x = 1120*sin(@)*cos(@) + 448*sin^2(@)

To find the max x we need to vary this with respect to @. So take the first derivative of x with respect to @ and set it to 0:
0 = 1120*cos^2(@) - 1120*sin^2(@) + 2*448*sin(@)*cos(@)

Analytically speaking, this is complicated to solve. If you are allowed, I would recommend you use a calculator to solve this. Otherwise I would start by noting that @ is in the first quadrant, so we may put:
cos(@) = +sqrt(1 - sin^2(@))

So the equation is:
0 = 1120*(1 - sin^2(@)) - 1120*sin^2(@) + 896*sin(@)*sqrt(1 - sin^2(@))

2240*sin^2(@) - 1120 = 896*sin(@)*sqrt(1 - sin^2(@)) <-- Square both sides.

(2240*sin^2(@) - 1120)^2 = 802816*sin^2(@)*(1 - sin^2(@))

5017600*sin^4(@) - 5017600*sin^2(@) + 1254400 = 802816*sin^2(@) - 802816*sin^4(@)

5820416*sin^4(@) - 5820416*sin^2(@) + 1254400 = 0

To make this look a little simpler, set z = sin^2(@):
5820416*z^2 - 5820416*z + 1254400 = 0

z^2 - z + 0.215517 = 0

z = (1 (+/-) sqrt((-1)^2 - 4*1*0.215517))/(2*1)

z = 0.685695 or z = 0.314305

So
sin(@) = (+/-)sqrt(z)
sin(@) = 0.82867 or sin(@) = 0.560629 <-- I'm only choosing the + solutions since @ is in QI.

Thus @ = 55.9007 degrees or @ = 34.0993 degrees.

As to what the plot looks like, look at the x equation again:
x = 70t*cos(@) + 1.75t^2

@ is just a constant, so this will be a parabola on a t vs. x plot.

-Dan
Hi once more Dan,

So I've spent some time looking through this. I don't know how you arrived with the 8 questions you came up with; further, I still don't understand the following:

You have the catapult WEST of the X (due west, like the problem states), but then why do you have it facing WEST? It should face EAST because there is a WESTERLY component. Does this change the whole problem? Ugh.

Many thanks!

9. Originally Posted by flash101
Hi once more Dan,

So I've spent some time looking through this. I don't know how you arrived with the 8 questions you came up with; further, I still don't understand the following:

You have the catapult WEST of the X (due west, like the problem states), but then why do you have it facing WEST? It should face EAST because there is a WESTERLY component. Does this change the whole problem? Ugh.

Many thanks!

Sorry, one more thing Dan.

I just graphed it (the y component of acceleration) and saw when it was highest and got 34 degrees, which was consistent for ONE of the angles you got..why do you have 2 angles? Graphing helped me avoid some of the tedious math. Hopefully getting the 34 degree angle wasn't just coincidence. You afterall got 2 angles.

10. I keep saying one more thing but something pops into my head as I submit! I got the max. distance as 380 meters; does this sound reasonable. If so, then I think we're on the same page...except with the additional angle.

11. Originally Posted by flash101
Hi once more Dan,
You have the catapult WEST of the X (due west, like the problem states), but then why do you have it facing WEST? It should face EAST because there is a WESTERLY component. Does this change the whole problem? Ugh.
The acceleration has a component in the Westerly directin so we would want the target to the West of the catapult. Now, it is possible with this acceleration to have the catapult facing EAST and hitting a target to the West, but this would not give us the maximum range. So we want the catapult to face West also.

Originally Posted by flash101
So I've spent some time looking through this. I don't know how you arrived with the 8 questions
Using the fact that the acceleration is constant we may simply integrate once and twice to get the equations:
v(t) = v0 + a*t
r(t) = r0 + v0*t + (1/2)a*t^2
where v(t) is the velocity vector, r(t) is the position vector (displacement), v0 is the initial velocity vector, and a is the acceleration vector. The two equations I used in the derivation of the solution are merely the equations for the components of r(t). Obviously we also get two equations for the components of the velocity vector. This accounts for 4 of the 8 equations I listed and are probably the only equations a Math text would list.

The other4 equations are a bit trickier to derive and I will merely state that their derivations can be found in any Introductory level Physics text. (It's the middle of the night, I don't have the derivations memorized, and LaTeX is down.) Since you aren't using them in solving the problem at hand I'm not going to worry about them.

-Dan

12. Originally Posted by flash101
Sorry, one more thing Dan.

I just graphed it (the y component of acceleration) and saw when it was highest and got 34 degrees, which was consistent for ONE of the angles you got..why do you have 2 angles? Graphing helped me avoid some of the tedious math. Hopefully getting the 34 degree angle wasn't just coincidence. You afterall got 2 angles.
It WOULD seem that to get the maximum range there shoudl only be one angle, shouldn't it? The only thing I can think of to suggest is to do a parametric plot of the trajectory for both angles and see what it looks like. (I'm not going to do it right now, but I may get to it later.)

However, obtaining two angles to hit a target isn't so strange. In our world (where the acceleration is only downward) when a projectile is fired to hit a target there generally is two angles you can use: the sum of these angles is 90 degrees. The projection angle for maximum range is singular: 45 degrees. But you have an accelertion component in the x direction as well, and that might add an additional solution.

It is also possible that my derivation managed to include an extra (incorrect) solution, as some solutions do. The quadratic equation was used, after all, and not all solutions from the quadratic equation are physically sensible. As I mentioned I didn't make the actual plots of the trajectory to verify both solutions would work. (Which is an admission of a failing on my part! )

-Dan