please help

[Noahb]

A particle is moving along the curve y=square root x. As the particle passes through the point (4,2), its x-coordinate increases at a rate of 3cm/s. How fast is the distance from the particle to the origin changing at this instant?

I Let z be the distance from the origin.
using the Pythagorean Theorem i found out z to be 4.5
i also found the derivative y'=1/2(X)^(-1/2)
but what exactly do i need to do to find what they are asking for?

thank you.

 

[pka]

\(\displaystyle \L
\begin{array}{l}
D = \sqrt {\left( {x - 0} \right)^2 + \left( {\sqrt x - 0} \right)^2 } = \sqrt {x^2 + x} \\
\frac{{dD}}{{dt}} = \frac{{2x + 1}}{{2 \sqrt {x^2 + x} }}\frac{{dx}}{{dt}}\quad ,\quad \left( {\frac{{dx}}{{dt}} = 3\quad \& \quad x = 4} \right) \\
\end{array}\)

 

[galactus]

Let Z=the distance from the point ot the origin.

Use Pythagoras:

\(\displaystyle Z^{2}=x^{2}+y^{2}\)

But \(\displaystyle y=\sqrt{x}\)

We have:

\(\displaystyle Z^{2}=x^{2}+x\) [1]

Also, \(\displaystyle Z=\sqrt{(4-0)^{2}+(3-0)^{2}}=2\sqrt{5}\)

differentiate [1] with respect to t:

\(\displaystyle \L\\2Z\frac{dZ}{dt}=2x\frac{dx}{dt}+\frac{dx}{dt}\)

\(\displaystyle \L\\\frac{dx}{dt}=3\) and \(\displaystyle \L\\x=4\)

Solve for \(\displaystyle \frac{dZ}{dt}\), you have all the values now.

 

[Noahb]

thanks

thank you so much for all the help..

 

[mike123]

would the final answer for the following problem when values are plugged in would be 13.5cm/s

dz/dt=2x(dx/dt)+(dx/dt) / 2
dz/dt= 2(4)(3)+(3) / 2
= 24+3/ 2
= 27 / 2
= 13.5

 

[jolly]

would the final answer for the following problem when values are plugged in would be 13.5cm/s

dz/dt=2x(dx/dt)+(dx/dt) / 2
dz/dt= 2(4)(3)+(3) / 2
= 24+3/ 2
= 27 / 2
= 13.5

Didn't you already say you weren't Noahb?