More Kinematics

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 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?

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 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)

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)

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)

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?

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)

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)

Originally posted by Aetherael
i think this mostly went to hell when i used the wrong initial equation y^2 = 2x haha

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.

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.....🙂

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/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!

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!

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

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 🙁