Important question please sove it for me!!!!!!!!!!!!!!!!!!?

Question

a particle is moving along the cure y = x ^ 1/2 ,where y and x are measured in cm.as the particle passes through the point (4,2).its x-coordinate is incearsing art rate of 3 cm/s.how fast the particle`s distance from the orgin changing as this instance?

Answer 1

You must compute the slope at which the particle is moving at
that point. Then the ratio of the absolute speed to the horizontal
speed is equal to the ratio of the hypotenuse of the slope to adjacent side. And of course the slope is computed by the derivative's value at x=4. So
y=x^1/2
dy/dx = 1/2(x^(-1/2)= 1/(2x^1/2)
dy/dx=1/4 at x=4 So the ratio of the vertical speed to the horizontal speed is 1/4 and the proportional hypotenuse in the direction of the particle(i.,e. the curve) is
(1^2 + 4^2)^1/2 = 17^1/2

So (17^(1/2))/4 = v/3cm/s where v is the velocity in direction
of the curve is

v=(3/4)(17^(1/2))

Stuart's approach is slightly different and fascinating to me
but also right.

Garipopki is also right but a bit work intensive I would say.

Garipopki's answer is closer to mine and I'm backing off
from supporting stuart. Work intensive or not, correctess
supercedes all.


At this point I'm responding to garypopki's comment that I'm
computing the velocity along the curve and not the velocity
away from thr origin. Good point. So the velocity away from
the origin is cos of the angle between the two directions times
the absolute velocity of the particle which is in the direction of the curve.
Angle of the curve at the point (4,2) is α where
tanα = 1/4, so α=14d (degrees, this is physics so I won't use
radians)

and the angle from the origin from which the particle is receding
is β where tanβ = 1/2, so β=26d

26d-14d=12d

So the component of velocity away from the origin is the absolute velocity times the cos of the angle between them.

Absolute velocity is 3.0923 so velocity from the origin
is cos(12)3.0923 =
3.024cm/s

which I have to admit is
awfully close to the 3.018 that you'd get from garypopki's
(27/20)5^(1/2)

Answer 2

I will try to sove this for you.

What is the particle's distance from the origin?

D = sqrt(x^2 + y^2)

Since x = y^2, we have

D = sqrt(x^2 + x)

Let's take the first derivative of this.

dD/dx = (1/2) * (x^2 + x)^-1/2 * (2x + 1)

At x = 4,

dD/dx = [(1/2) / sqrt(20)] * 9
= (9/2) / 2sqrt(5)
= 9/4 / sqrt(5)
= 9sqrt(5) / 20

Now, dx/dt is given as 3, so

dD/dt = 27sqrt(5) / 20

gopal differentiated wrong, and then forgot about the 3 cm/sec, which is nowhere to be found is his solution.

Albert shows that the .25 Stuart got is not the velocity in the y-direction (Stuart says acceleration but means velocity), but the ratio of the x- and y-velocities. Stuart arrives at .25 without using the 3 cm/sec, so cannot be correct. Wisely, Albert does not support Stuart's reasoning. Stuart's answer of 3.01 is too small becuase he used .25 as the y-velocity instead of using 1/4 of the x-velocity, as Albert did, who arrived at 3.09.

BUT!!! Stuart and Albert are both attempting to compute the velocity IN THE DIRECTION OF THE CURVE, which is NOT WHAT IS ASKED FOR. We are asked for the rate of change of distance between the particle and the origin, and only my method and Gopal's provide the answer, but Gopal messed up the computations.

Answer 3

Differentiate y=x^1/2 to get y=1/2x^(-1/2)

Take the coordinate x=4 and plug it in to find the rate of acceleration in the y direction is 1/2*4^(-1/2) = 0.25 cm/s

Use Pythagoras to combine the vectors of acceleration 3 and 0.25: (3^2+0.25^2)^1/2 = sqrt(9.0625) cm/s

This is the answer (I think) but wait for someone to verify cos I might be wrong.

Answer 4

y=x^1/2
s=rt(x^2+y^2)
s=rt(x^2+x)
ds/dt=[1/2(x^2+x)^1/2]*(2x+1)
=9/2rt12
=9/4rt3 cm /sec