How fast is the distance from the particle to the origin changing at this instant?

Question

the particle is moving along the curve y=sqrtx. 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 don't know how to solve the question.  I know it has something to do with tangent lines and derivatives but not sure how to start.

Answer 1

Sounds like a related rate problem.  No tangent lines needed.  Your particle travels along a curve, and its distance from the origin is given by that of a straight line from the origin to the particle.  After placing the particle on an arbitrary point on the curve y=√(x), drop a perpendicular from the particle to the x-axis, and then draw a line from the origin to the particle.  You'll see a right triangle.  The length of the horizontal leg is x, and the length of the vertical leg is y.  I encourage you to draw this, as it completely summarizes the situation graphically.

 

Let the distance from the origin to the particle be D.  By the Pythagorean Theorem (as you'll see plainly if you sketch it out), D = (x² + y²)^1/2.  Since y = √(x), this comes out to D = (x² + x)^1/2.  I'm using rational exponents rather than roots to aid in using the power rule for derivatives, which must be applied next.

 

Now, differentiate both sides of the equation with respect to time.  You'll need to use the chain rule on the right hand side:

 

dD/dt = 1/2(x² + x)^-1/2 * (2x+1) * (dx/dt).  Let's simplify a little bit:

 

dD/dt = (2x+1)/[2√(x²+x)] * (dx/dt).

 

From the problem statement, we know that x = 4 cm, and dx/dt = 3 cm/s.  Plug these into the above equation and you'll find the rate at which the particle's distance from the origin is changing with respect to time.