1. ## Parametric Equations

I need help with part(a). I can find the parametric equations for a nonmoving ladder, but when the ladder begins to move i'm having difficulty.

http://www2.palomar.edu/users/canfin...0Equations.pdf

2. Originally Posted by rico3639
I need help with part(a). I can find the parametric equations for a nonmoving ladder, but when the ladder begins to move i'm having difficulty.

http://www2.palomar.edu/users/canfin...0Equations.pdf
at time 't' the point A is at (3t,0). let the point B be at (0,y(t)) at time 't'. Then $9t^2+y^2(t)=12^2.$ why?
this will give you the coordinated of point B with time. Find the coordinates of M using the above. now the monkey is at a distance of $2t$ from M.(why?)
can you now locate the coordinates of the monkey at time 't'. That when done will give you the required parametric equation.

3. Originally Posted by abhishekkgp
at time 't' the point A is at (3t,0). let the point B be at (0,y(t)) at time 't'. Then $9t^2+y^2(t)=12^2.$ why?
this will give you the coordinated of point B with time. Find the coordinates of M using the above. now the monkey is at a distance of $2t$ from M.(why?)
can you now locate the coordinates of the monkey at time 't'. That when done will give you the required parametric equation.
Not understanding how you get (3t,0) at point A. What is the "coordinated" of point B with time mean?
Not sure how you get the monkey at a distance of 2t from M.

4. Originally Posted by rico3639
Not understanding how you get (3t,0) at point A. What is the "coordinated" of point B with time mean?
Not sure how you get the monkey at a distance of 2t from M.
Ok I see why you chose 3t,0 and I also see why you chose (0, y(t)). Please explain the monkey part some more. It's at a distance of 2t, but how do I use this? Please help.

5. Originally Posted by rico3639
Ok I see why you chose 3t,0 and I also see why you chose (0, y(t)). Please explain the monkey part some more. It's at a distance of 2t, but how do I use this? Please help.
The coordinates of A at time 't' will be $(3t,0)$
Let the coordinates of B at time 't' be $(0,b(t))$(i am purposely not using y(t) you will see why.)

The equation of the line AB at time 't' is given by :

$\frac{y-b(t)}{x-0}=\frac{b(t)-0}{0-3t}$.(do you see why?)
The coordinates of M at time 't' are:
$(\frac{3t}{2},\frac{b(t)}{2})$(why?)
The monkey is at a distance of 2t from M so the monkey will be found somewhere on the circle centred at M and of radius 2t.(make sure you understand why this is so). At the same time the monkey is also on the ladder,that is, the monkey will also be somewhere on line AB. The intersection of the circle (mentioned above in green) and AB will give you the position of the monkey.
The equation of that circle is ${(x-\frac{3t}{2})}^2+{(y-\frac{b(t)}{2})}^2=(2t)^2$.
This will intersect AB at two points. One of these is the position of the monkey. Can you figure out which one.

6. Originally Posted by abhishekkgp
The coordinates of A at time 't' will be $(3t,0)$
Let the coordinates of B at time 't' be $(0,b(t))$(i am purposely not using y(t) you will see why.)

The equation of the line AB at time 't' is given by :

$\frac{y-b(t)}{x-0}=\frac{b(t)-0}{0-3t}$.(do you see why?)
The coordinates of M at time 't' are:
$(\frac{3t}{2},\frac{b(t)}{2})$(why?)
The monkey is at a distance of 2t from M so the monkey will be found somewhere on the circle centred at M and of radius 2t.(make sure you understand why this is so). At the same time the monkey is also on the ladder,that is, the monkey will also be somewhere on line AB. The intersection of the circle (mentioned above in green) and AB will give you the position of the monkey.
The equation of that circle is ${(x-\frac{3t}{2})}^2+{(y-\frac{b(t)}{2})}^2=(2t)^2$.
This will intersect AB at two points. One of these is the position of the monkey. Can you figure out which one.
When you say that the coordinates of M are located at 3t/2, b(t)/2, are you implying that the monkey is halfway on the ladder? The monkey is not halfway on the ladder.

7. Originally Posted by rico3639
When you say that the coordinates of M are located at 3t/2, b(t)/2, are you implying that the monkey is halfway on the ladder? The monkey is not halfway on the ladder.
M is not where the monkey is. M FOR "MID POINT".

8. On the problem it says "Suppose a monkey is at a point M on the ladder"=/

9. Originally Posted by rico3639
When you say that the coordinates of M are located at 3t/2, b(t)/2, are you implying that the monkey is halfway on the ladder? The monkey is not halfway on the ladder.
i am so sorry i misread the problem. i took it as "monkey starts from the mid point of the ladder" but actually it says that the monkey starts from A. If you wait for sometime i can re-post the correct solution.

10. Thanks a lot. I'll continue to work on it, hopefully I can check with your solution soon.

11. With slight changes the correct solution is:

The coordinates of A at time 't' will be $(3t,0)$
Let the coordinates of B at time 't' be $(0,b(t))$(i am purposely not using y(t) you will see why.)

The equation of the line AB at time 't' is given by :

$\frac{y-b(t)}{x-0}=\frac{b(t)-0}{0-3t}$.(do you see why?)

The monkey is at a distance of 2t from A so the monkey will be found somewhere on the circle centred at A and of radius 2t.(make sure you understand why this is so). At the same time the monkey is also on the ladder,that is, the monkey will also be somewhere on line AB. The intersection of the circle (mentioned above in green) and AB will give you the position of the monkey.
The equation of that circle is $(x-3t)^2+(y-0)^2=(2t)^2$.
This will intersect AB at two points. One of these is the position of the monkey. Can you figure out which one?

12. in my previous post you will need to find $b(t)$ using $b^2(t)+(3t)^2=(12)^2$

13. Originally Posted by abhishekkgp
in my previous post you will need to find $b(t)$ using $b^2(t)+(3t)^2=(12)^2$
Once I find b(t) using that equation, do I plug b(t) into the y= -b(t)x/3t + b(t) equation to get y in terms of x and t?

I'm a little confused on what I do next. Now that b(t) is not really y, where am I suppose to get my 2 parametric equations from?

I have y = (- (144-9t^2)^.5 )*x/(3t) + (144 - 9t^2)^.5

I took b(t) and plugged it into the equation of the line AB.

14. My concern as follows:

(i) The equation of the circle and the equation of the line AB need to intersect. Which equations do I use in order to accomplish this?

15. Originally Posted by rico3639
My concern as follows:

(i) The equation of the circle and the equation of the line AB need to intersect. Which equations do I use in order to accomplish this?
to find the intersection of the circle and the line you of course need their corresponding equations.
the equations of them are :
1)Line AB: $\frac{y-b(t)}{x-0}=\frac{b(t)-0}{0-3t}$. (plug in b(t) you have found out. i would advice that you find the coordinates of Monkey(M) in terms of b(t) and then plug in the b(t) you have found out.)

2) circle: $(x-3t)^2+(y-0)^2=(2t)^2$

solve the above two equations simultaneously to get x= some whacky expression in 't'; y= another whacky expression in 't'. these are the coordinates of the monkey.

now is it clear ?

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