- 22 Jan '10 03:45 / 6 editshello all, stuck on a problem looking for some insight

A particle travels along the path, y^2 = 4x with a constant speed of V = 4(m/s). determine the x, and y components of the particles velocity and acceleration when the particle is at x = 4(m)

im going to use prime notation for ease of formatting

heres where im at so far

a particles velocity is tangent to its path so I need to find the slope of the function at the given x

y^2 = 4x

2y*y'(x) = 4

y'(x) = 2/(y)

y'(4) = 1/2

from here I can find Vx,and Vy (the x-y components of the velocity respectivley)

from the right triangle (2,1,Sqrt(5))

@x=4......{y'(t) =Vy = ||V||*1/sqrt(5) = 4/sqrt(5)

..............{x'(t) = Vx = ||V||*2/sqrt(5) = 8/sqrt(5)

Now to find the components of the acceleration

y'(x) = 2/y

y'(t) = (2/y)(x'(t))

y"(t) = 2*[(-1/y^2)*(y'(t))*(x'(t)) + (1/y)(x"(t))]

here is where im stuck

1 equation 2 unknows and i'm having trouble coming up with another independent equation for either

y"(t) or x"(t)

does anyone see something about the acceleration that could help me out? - 22 Jan '10 08:24well you can probably make your life a little easier by using y^2 = 2x to see that y = (2x)^1/2

so y'(t) = 2/y * (x'(t))= 2/(2x)^1/2 * (8/sqrt(5))= 8/5*sqrt(10)*x^(-1/2) -- note i hate doing algebra in text i have no idea if i messed up that simplification

now y''(t) = [(-1/2)*(8/5)*(sqrt(10))x^(-3/2)] * x'(t)??? ... though this is why i ALSO hate prime notation: i'm taking t-partials in an equation with x's so do we need the chain rule there?

if this is accurate y''(t) = [(-1/2)*(8/5)*(sqrt(10))x^(-3/2)] * 8/sqrt(5) = -64/10*sqrt(2) * x^(-3/2) - 22 Jan '10 08:58

Why so difficult?*Originally posted by Aetherael***well you can probably make your life a little easier by using y^2 = 2x to see that y = (2x)^1/2**

so y'(t) = 2/y * (x'(t))= 2/(2x)^1/2 * (8/sqrt(5))= 8/5*sqrt(10)*x^(-1/2) -- note i hate doing algebra in text i have no idea if i messed up that simplification

now y''(t) = [(-1/2)*(8/5)*(sqrt(10))x^(-3/2)] * x'(t)??? ... though this is why i ALSO hat ...[text shortened]... rate y''(t) = [(-1/2)*(8/5)*(sqrt(10))x^(-3/2)] * 8/sqrt(5) = -64/10*sqrt(2) * x^(-3/2)

If y^2 = 4x

Then y = -2*sqrt(x) or y = 2*sqrt(x)

Then y'= -1/sqrt(x) or y'= 1/sqrt(x)

Then y'' = 1/2x*sqrt(x) or y'' = -1/2x*sqrt(x) - 22 Jan '10 13:36

Yeah, the chain/product rule is required, and unfortunatley that is not the given solution. the given is*Originally posted by Aetherael***well you can probably make your life a little easier by using y^2 = 2x to see that y = (2x)^1/2**

so y'(t) = 2/y * (x'(t))= 2/(2x)^1/2 * (8/sqrt(5))= 8/5*sqrt(10)*x^(-1/2) -- note i hate doing algebra in text i have no idea if i messed up that simplification

now y''(t) = [(-1/2)*(8/5)*(sqrt(10))x^(-3/2)] * x'(t)??? ... though this is why i ALSO hat ...[text shortened]... rate y''(t) = [(-1/2)*(8/5)*(sqrt(10))x^(-3/2)] * 8/sqrt(5) = -64/10*sqrt(2) * x^(-3/2)

Ax = .32(m/s^2)

Ay = -.64(m/s^2)

So how to get there? haha - 22 Jan '10 16:32

Pushed for time but my initial thought is to explicitly find the x component of acceleration to get x''(t) .. just as you did for y''(t).*Originally posted by joe shmo***hello all, stuck on a problem looking for some insight**

A particle travels along the path, y^2 = 4x with a constant speed of V = 4(m/s). determine the x, and y components of the particles velocity and acceleration when the particle is at x = 4(m)

im going to use prime notation for ease of formatting

heres where im at so far

a particles velocity ...[text shortened]... x"(t)

does anyone see something about the acceleration that could help me out?

That should give you second equation. - 22 Jan '10 17:52 / 2 edits

but your y' and y'' here are dy/*Originally posted by TheMaster37***Why so difficult?**

If y^2 = 4x

Then y = -2*sqrt(x) or y = 2*sqrt(x)

Then y'= -1/sqrt(x) or y'= 1/sqrt(x)

Then y'' = 1/2x*sqrt(x) or y'' = -1/2x*sqrt(x)**dx**and dy^2/d^2x which are derivatives of y with respect to x... but the problem isn't concerned with that relationship so much as how each y and x are changing with respect to a time parameter. maybe only implicitly (in discussion of velocity and acceleration) but still since we need acceleration and velocity x-partials and y-partials i thought it might be useful.

that being said, your method is probably useful for finding the TOTAL acceleration, and then finding which parts of it are in which directions using the formulas joe used in his post above to find the Vx and Vy - 22 Jan '10 18:06

i think this mostly went to hell when i used the wrong initial equation y^2 = 2x haha*Originally posted by Aetherael***well you can probably make your life a little easier by using y^2 = 2x to see that y = (2x)^1/2**

so y'(t) = 2/y * (x'(t))= 2/(2x)^1/2 * (8/sqrt(5))= 8/5*sqrt(10)*x^(-1/2) -- note i hate doing algebra in text i have no idea if i messed up that simplification

now y''(t) = [(-1/2)*(8/5)*(sqrt(10))x^(-3/2)] * x'(t)??? ... though this is why i ALSO hat ...[text shortened]... rate y''(t) = [(-1/2)*(8/5)*(sqrt(10))x^(-3/2)] * 8/sqrt(5) = -64/10*sqrt(2) * x^(-3/2) - 22 Jan '10 21:23

Fret no more, I figured it out my other piece of information.*Originally posted by Aetherael***i think this mostly went to hell when i used the wrong initial equation y^2 = 2x haha**

It turns out that since the velocity is constant that the sum of the accelerations in the direction of V must be zero

letting B be the angle between V and the x axis

Ax cos(B) = Ax:V (the component of the acceleration in the direction of V)

Ay sin(B) = Ay:V

Ax*2/sqrt(5) + Ay*1/sqrt(5) = 0

x"(t) = -y"(t)*1/2

sub it in and solve... - 22 Jan '10 23:09

Just out of cuiosity how would you approach that, because I tried to solve for x"(t) explicitly, but substitution yielded division by 0. There is a possibility I could have done it wrong, but maybe you have a different perspective? Since the cat is caught, I just want to see if there is more than one way to skin it so to speak.....*Originally posted by wolfgang59***Pushed for time but my initial thought is to explicitly find the x component of acceleration to get x''(t) .. just as you did for y''(t).**

That should give you second equation. - 23 Jan '10 10:38

Work in Progress ... long time since I did any calculus!!!!!*Originally posted by joe shmo***Just out of cuiosity how would you approach that, because I tried to solve for x"(t) explicitly, but substitution yielded division by 0. There is a possibility I could have done it wrong, but maybe you have a different perspective? Since the cat is caught, I just want to see if there is more than one way to skin it so to speak.....** - 07 Mar '10 22:14 / 4 editsHere's my working:

y^2 = 4x

so: y = 2sqrt(x)

differentiating

dy/dx = 1/sqrt(x)

rearranging we get dy = dx/sqrt(x)

similarly, we can write the first equation as:

x = y^2/4

and differentiate to get

dx/dy = y/2

which rearranges to

dx = y.dy/2

we know speed is 4 m/s

i.e. sqrt((dx/dt)^2 + (dy/dt)^2) = 4

squaring both sides: (dx/dt)^2 + (dy/dt)^2 = 16

substituting in the expression we got for dy:

(dx/dt)^2 + (dx/dt)^2/x = 16

((x - 1)/x)(dx/dt)^2 = 16

dx/dt = 4x/(1-x) {equation a}

using the quotient rule to differentiate equation a

ddx/dt^2 =4/(1-x)^2 {equation b}

going back to the equation:

(dx/dt)^2 + (dy/dt)^2 = 16

substituting in the equation we got for dx:

y^2.(dy/dt)^2/4 + (dy/dt)^2 = 16

5y^2/4(dy/dt)^2 = 16

(dy/dt)^2 = 4*16.y^(-2)/5

dy/dt = 8/(sqrt(5)y) {equation c}

differentiatng that

ddy/dt^2 = -8/(sqrt(5)y^2) {equation d}

and we get the necessary answers at {x= 4, y=4} by substituting those values into equations a,b,c and d

x = 4, y=4

dx/dt = -16/3

ddx/dt^2 = 4/9

dy/dt = 2/sqrt(5)

ddy/dt^2 = -1/(2*sqrt(5))

Hmm, I think I made a mistake somewhere, but tired now, going to bed and checking it over tomorrow! - 08 Mar '10 03:46

in your equation "a"*Originally posted by iamatiger***Here's my working:**

y^2 = 4x

so: y = 2sqrt(x)

differentiating

dy/dx = 1/sqrt(x)

rearranging we get dy = dx/sqrt(x)

similarly, we can write the first equation as:

x = y^2/4

and differentiate to get

dx/dy = y/2

which rearranges to

dx = y.dy/2

we know speed is 4 m/s

i.e. sqrt((dx/dt)^2 + (dy/dt)^2) = 4

squaring both sides: (dx/dt)^2 + (dy/d ...[text shortened]... I think I made a mistake somewhere, but tired now, going to bed and checking it over tomorrow!

You just slipped up on the algebra

should be

(dx/dt)^2 + (1/x)(dx/dt)^2 =16

(dx/dt)^2(1+x) = 16x

dx/dt = 4* sqrt(x/(1+x))

but other than that, thats the alternative I was looking for.

Thanks - 08 Mar '10 08:20 / 2 edits

Ah, thanks for that Joe.*Originally posted by joe shmo***in your equation "a"**

You just slipped up on the algebra

should be

(dx/dt)^2 + (1/x)(dx/dt)^2 =16

(dx/dt)^2(1+x) = 16x

dx/dt = 4* sqrt(x/(1+x))

but other than that, thats the alternative I was looking for.

Thanks

I think the long form of writing differentiation as dx/dy helps, because the dx etc can be used as normal variables and rearranged. I don't find the x'' notation very helpful usually. I'm still worried about my calculation though because the speed at that point doesn't seem to come out as 4 - 08 Mar '10 14:39

Just a slip in equation c*Originally posted by iamatiger***Ah, thanks for that Joe.**

I think the long form of writing differentiation as dx/dy helps, because the dx etc can be used as normal variables and rearranged. I don't find the x'' notation very helpful usually. I'm still worried about my calculation though because the speed at that point doesn't seem to come out as 4

1/4*y^2*(dy/dt)^2 + (dy/dt)^2 =16

(dy/dt)^2*(1/4*y^2 + 1) = 16

dy/dt = 8/sqrt(y^2 +4)

rough night?