another systems of equations problem

[allegansveritatem]

"g" is standard notation for the downward acceleration due to gravity and "sx" and "sy" are standard notation for x and y coordinates of position. However, this problem is giving you a different notation so you don't want to use those. In this problem the position of the ball is given by x and y and the acceleration is incorporated into "a".

In this problem, you are told that "using Calculus it can be shown that the path of the ball is given by y= ax^2+ x+ c for some constants a and c". So I take it you have not yet taken Calculus and are not expected to derive that. Since "real life" problems do not have a coordinate system attached you do need to select one. It is standard practice to have the x-axis horizontal and the y-axis vertical and the equation "y= ax^2+ x+ c" is assuming that. You do have to decide where to put the origin of the coordinate system. I see two points where it would be reasonable to set the origin- at the initial point of the trajectory or and the end point. Choosing that initial point as the origin, x= 0 and y= 0 we must have y= 0= a(0^2)+ b(0)+c= c. So now we know the equation is y= ax^2+ x. The end point is 50 m down the hill which has a "slope of -3/4". The slope (not quite Calculus) is "y/x= -3/4" so y= -3x/4. Using the Pythagorean formula, x^2+ y^2= x^2+(3/4)^2x^2=(1+9/16)x^2= (25/16)x^2= 50^2 so x^2= (50^2)(16/25)= (2)(50)(16)=(4)(25)(16). Taking the square root of both sides, x= (2)(5)(4)= 40. y=-3x/4= -30. In coordinate system with origin at the starting point the end point has coordinates x= 40, y= -30. Putting those into y= ax^2+ x, -30=a(40)^2+ 40. -70= 1600a so a= -7/160.

I think I have pretty much done what you point to here, no?

 

[Steven G]

I now realize that you posted this under algebra, not calculus. Sorry about that.

You know that y=ax^2 + x + c (given).

So there are two unknowns, namely a and c. So you need two equations in a and c.

If you place the origin, (0,0) at the start of the projectile that will give you 0 = a*0^2 + 0 + c.

Now you can figure out the endpoint of the trajectory. Not to just give you the 2nd equation, let's call that endpoint (r, s).
Then the 2nd equation will be s = a*r^2 + r + c

Solve the two equations and move on. Sorry for making this harder than necessary.

 

[jonah2.0]

Dull sober goggles contribution follows.

Here what I did for the vertex. No?:
View attachment 21348
I have looked at the Swokowski book section 2.6 long and hard in reviewing parabolas and vertices.

Close but no cigar.
Take another look. Subconscious assimilation will kick in sooner or later and you're bound to experience that eureka moment anytime soon.
In the meantime, let me paint you a picture.

 

[allegansveritatem]

Dull sober goggles contribution follows.

Close but no cigar.
Take another look. Subconscious assimilation will kick in sooner or later and you're bound to experience that eureka moment anytime soon.
In the meantime, let me paint you a picture.
View attachment 21351View attachment 21352

OK. I see that and that information is on the little diagram that I drew in my last post. I found x =40 and y=-30 from the dimensions of the triangle. But they are asking for a b and c. C is easy. Because of the 0,0 point we know c is 0. There remains a and b. Seems to me I need to know another point on the parabola to devise a system that involves a and b. Using the point 0,0 does not work, at least I couldn't find a way with it. Using 0,0 one gets an a that is zero. I have a haunting feeling that there is something obvious that I am missing in all this but, as you say, until the mist clears
the path is lost. I will work on this again today and see what turns up.

 

[allegansveritatem]

I now realize that you posted this under algebra, not calculus. Sorry about that.

You know that y=ax^2 + x + c (given).

So there are two unknowns, namely a and c. So you need two equations in a and c.

If you place the origin, (0,0) at the start of the projectile that will give you 0 = a*0^2 + 0 + c.

Now you can figure out the endpoint of the trajectory. Not to just give you the 2nd equation, let's call that endpoint (r, s).
Then the 2nd equation will be s = a*r^2 + r + c

Solve the two equations and move on. Sorry for making this harder than necessary.

Yes, I am going to go back at this today and see what I can do. I know two points but I think I need to know another point because I am having a hard time building a system with that 0,0 point. I will see what comes today.

 

[Steven G]

Yes, I am going to go back at this today and see what I can do. I know two points but I think I need to know another point because I am having a hard time building a system with that 0,0 point. I will see what comes today.

I gave you the system!
The origin becomes 0 = a*0^2 + 0 + c which means c=0. Is that clear?

 

[allegansveritatem]

Dull sober goggles contribution follows.

Close but no cigar.
Take another look. Subconscious assimilation will kick in sooner or later and you're bound to experience that eureka moment anytime soon.
In the meantime, let me paint you a picture.
View attachment 21351View attachment 21352

can you tell me how this that I have done is any different from what Hallsofivy posted in #12? I don't see it:

 

[Steven G]

What happened to c, where did b come from and why is a(1^2) + b(1) = 1?

 

[jonah2.0]

Dull sober goggles hints and giveaways follow.

can you tell me how this that I have done is any different from what Hallsofivy posted in #12? I don't see it:
View attachment 21364

I see your efforts.
HallsofIvy already explained it well.
Here's my take on it:
From the diagram, we know the first point of intersection is (0, 0).
Thus
y = a*x^2 + x + c
0 = a*0^2 +0 + c
c = 0
and
y = a*x^2 + x + 0 or
y = a*x^2 + x
We now have
y = a*x^2 + x
and
y = -3/4*x
With -3/4*x = a*x^2 + x
We get a = -7/(4*x)
Applying the Pythagorean Theorem for the 2nd point of intersection, with
x^2 + y^2 = 50^2 and y = -3/4*x
We have
x^2 + (-3/4*x)^2 = 50^2
Clearly, the positive value of x will be "it" and plugging that in the derived value of "a" should give you the value of "a" for y = a*x^2 + x
You then need a review of "completing the square" to determine the maximum height of the ball off the ground.

 

[jonah2.0]

Dull sober goggles hints and giveaways follow.

I see your efforts.
HallsofIvy already explained it well.
Here's my take on it:
From the diagram, we know the first point of intersection is (0, 0).
Thus
y = a*x^2 + x + c
0 = a*0^2 +0 + c
c = 0
and
y = a*x^2 + x + 0 or
y = a*x^2 + x
We now have
y = a*x^2 + x
and
y = -3/4*x
With -3/4*x = a*x^2 + x
We get a = -7/(4*x)
Applying the Pythagorean Theorem for the 2nd point of intersection, with
x^2 + y^2 = 50^2 and y = -3/4*x
We have
x^2 + (-3/4*x)^2 = 50^2
Clearly, the positive value of x will be "it" and plugging that in the derived value of "a" should give you the value of "a" for y = a*x^2 + x
You then need a review of "completing the square" to determine the maximum height of the ball off the ground.
View attachment 21372

View attachment 21373View attachment 21374View attachment 21375

Alternatively, you may want to consider what Sir Subhotosh Khan said, post #17 (once you finally understand how the value of a is derived and how you arrive at the parabola function):
The vertex will be located at the mid-point of these roots. The y-coordinate of the vertex is the maximum height (off the x-axis) attained by the projectile.
Also,

 

[allegansveritatem]

I gave you the system!
The origin becomes 0 = a*0^2 + 0 + c which means c=0. Is that clear?

yes, I have already used that and know that c =zero. I have also, found that one of the points on the parabola is 40, -30. So I know two points 0,0 and 40, -30. Now, 0,0 is not suitable to build a system with using a and b. So I, seeing that the ball starts out on a 45 degree angle took as one of the points I could use to build a system. 1,1. So Here is what I did. What is wrong with it?:

and for b:

and for vertex:

Alternatively, you may want to consider what Sir Subhotosh Khan said, post #17 (once you finally understand how the value of a is derived and how you arrive at the parabola function):
The vertex will be located at the mid-point of these roots. The y-coordinate of the vertex is the maximum height (off the x-axis) attained by the projectile.
Also,View attachment 21379

The examples you quote, or something very like them , look wonderfully familiar to me, having, as I have recently, gone over them with a fine comb myself. But that is not to say that I couldn't revisit them with much profit, which I will. You seem to be saying that the way I derived the solution is wrong because it is based on conjecture? Or is it absolutely wrong, wrong in itself? I mean wrong like falling down bridge wrong. Anyway, I will study your post over later and work it out and see if I get a clearer idea of the situation. I shall return. Thanks for these pointers.

 

[Steven G]

I am confused as to why you can't use (0,0). Didn't you use it to find out c=0? Where did b come from?

Solution:
y = ax^2 + x + c
0=a*0^2 + 0 + c => 0 = c
-30 = a*40^2 + 40 + c => - 30 =1600a + 40 + c

The two bold equations above is the system you need to solve.

Easily we have c=0. Plugging that into the other equation we get -30 = 1600a + 40 =>-7 = 160a => a = -7/160
Solution:
y = ax^2 + x + c
0=a*0^2 + 0 + c => 0 = c
-30 = a*40^2 + 40 + c => - 30 =1600a + 40 + c

The two bold equations is the system you need to solve.

Easily we have c=0. Plugging that into the other equation we get -30 = 1600a + 40 =>-7 = 160a => a = -7/160

Solution: y = -7/60*x^2 + x

 

[allegansveritatem]

I am confused as to why you can't use (0,0). Didn't you use it to find out c=0? Where did b come from?

Solution:
y = ax^2 + x + c
0=a*0^2 + 0 + c => 0 = c
-30 = a*40^2 + 40 + c => - 30 =1600a + 40 + c

The two bold equations above is the system you need to solve.

Easily we have c=0. Plugging that into the other equation we get -30 = 1600a + 40 =>-7 = 160a => a = -7/160
Solution:
y = ax^2 + x + c
0=a*0^2 + 0 + c => 0 = c
-30 = a*40^2 + 40 + c => - 30 =1600a + 40 + c

The two bold equations is the system you need to solve.

Easily we have c=0. Plugging that into the other equation we get -30 = 1600a + 40 =>-7 = 160a => a = -7/160

Solution: y = -7/60*x^2 + x

Taking some hints from Jonah and others I came up with this:

and:

I tried to use completing the square for vertex but ran into a problem. What happened I can't say. I got the x factor but y went wild. I tried this several times and kept coming up with crazy stuff:Here is what happened:

 

[allegansveritatem]

I am confused as to why you can't use (0,0). Didn't you use it to find out c=0? Where did b come from?

Solution:
y = ax^2 + x + c
0=a*0^2 + 0 + c => 0 = c
-30 = a*40^2 + 40 + c => - 30 =1600a + 40 + c

The two bold equations above is the system you need to solve.

Easily we have c=0. Plugging that into the other equation we get -30 = 1600a + 40 =>-7 = 160a => a = -7/160
Solution:
y = ax^2 + x + c
0=a*0^2 + 0 + c => 0 = c
-30 = a*40^2 + 40 + c => - 30 =1600a + 40 + c

The two bold equations is the system you need to solve.

Easily we have c=0. Plugging that into the other equation we get -30 = 1600a + 40 =>-7 = 160a => a = -7/160

Solution: y = -7/60*x^2 + x

I will study over your post and reply to it in the morning. I can say now that I did use the 0,0 to get c, but when I tried to use it to build a system using a and b, I failed. I kept getting a=0.

 

[Steven G]

This is like the third time I am asking you what b is? The problem never had a b! If you think there should be a b, then that is fine but can you please define it? Do you recall that the original problem was to find a and c in y=a*x^2 + x + c. There is no b!

 

[jonah2.0]

Beer soaked comments follow.

You were all good when you arrived at [MATH]a=-\frac{7}{160}[/MATH]You should have stopped there and plugged that value of [MATH]a[/MATH] straighaway in [MATH]y=ax^{2}+x+c[/MATH]Condition of requirement (a) right there is fulfilled with [MATH]a=-\frac{7}{160}[/MATH] and c = 0
Didn't you see my 1st equation and its corresponding graph at post #23?
You just now need to complete the square (See Example 4, Post #29) for requirement (b).

Edit: I guess you did arrive at the vertex coordinates from your 2nd picture; too hammered right now to analyze how but the coordinates seem to be right.

Edit 2: Ah yes, you used the Theorem indicated at post #30

Edit 3: Good stuff. I guess you just need more practice with that completing the square thingamajig.

 

[jonah2.0]

Beer soaked contribution follows.

Something for you to ponder if you don't have GeoGebra CAS Calculator yet.

 

[Steven G]

Ah, the b is from the vertex form. Unless I missed it, I haven't seen your equation yet. y = ....

 

[allegansveritatem]

This is like the third time I am asking you what b is? The problem never had a b! If you think there should be a b, then that is fine but can you please define it? Do you recall that the original problem was to find a and c in y=a*x^2 + x + c. There is no b!

well, b is 1 is it not? I think I showed that in my last post. Wait,let me check that post. OK. I see that I didn't post the image where I found that b is one. I have that on another computer and will post it. That was easy to find once I got a and c. As for the system you suggest, it just never occurred to me to do it that way. I was intent on turning a(0)+b(0)=0 and a(40^2) +b(40)--30 into a system and couldn't figure it out. I kept getting 0 as a coefficient for a somehow. Maybe it was just impatience on my part. Anyway, I will try your approach just to see how it works. Thanks

 

[allegansveritatem]

Beer soaked comments follow.

You were all good when you arrived at [MATH]a=-\frac{7}{160}[/MATH]You should have stopped there and plugged that value of [MATH]a[/MATH] straighaway in [MATH]y=ax^{2}+x+c[/MATH]Condition of requirement (a) right there is fulfilled with [MATH]a=-\frac{7}{160}[/MATH] and c = 0
Didn't you see my 1st equation and its corresponding graph at post #23?
You just now need to complete the square (See Example 4, Post #29) for requirement (b).

Edit: I guess you did arrive at the vertex coordinates from your 2nd picture; too hammered right now to analyze how but the coordinates seem to be right.

Edit 2: Ah yes, you used the Theorem indicated at post #30

Edit 3: Good stuff. I guess you just need more practice with that completing the square thingamajig.

No, I actually did do what you suggest above. Here is the image. I thought I had uploaded it yesterday but I did not. Here is what I did for b:

I didn't not even copy this last night from my camera for some reason but I found it on the sd card just now. That b part was easy. I only had one variable to worry about. As for the completing the square, I will go back on that today and find what I did wrong.