<p>For the masters of calculus out there, I am having trouble with these two homework problems from the book. I anyone can help me with either of them, especially the first one, it would be great.</p>
<p>A particle is moving along the curve y=sqroot(x). As the particle passes through point (4,2), its x-coordinate increases at a rate of 3 cm/s. How fast is the distance from the particle to the origin changing at this instant?</p>
<p>A trough is 10 ft long and its ends have the shape o isoscelestriangles that are 3 ft across at the top and have height of 1ft. If the trough is being filed with water at a rate of 12 ft^3/min, how fast is the water level rising when the water is 6 in deep?</p>
<p>I know it is tough to show the steps without a drawing, but any help is much appreciated.</p>
<p>ok so d = square root of x^2 + y^2 so you gotta take the derivative of that (write y = sqrt of x as a substitution) with respect to time, and then u should get an expression with x's and a dx/dt and then you just plug in the numbers they give and you should get your answer (because remember, you want to find the rate of change of the total distance, not the x-position)</p>
<p>I hate water level problems so I'll leave someone else to do that one lol.</p>
<p>^thanks piccolo. yeah, that water problem is a pain. Does anyone know how to do it?</p>
<p>Okay, the second one is just a related rate problem.</p>
<p>You are given the information for dV and h. Thus this is a "hint" that you want to formulate an expression that relates V to h and differentiate it.</p>
<p>You should see that V = 15h^2</p>
<p>Differentiate this, plug in numbers, and solve for h.</p>
<p>piccolo, when I take the derivative of x^2+ (sqroot x)^2 = d^2,
I end up with two dx/dt . Is this normal, should I plug in twice?</p>
<p>Actually, I can factor the dx/dt, so I guess it works...</p>
<p>il bandito(or anyone),if it is not too much trouble, can you briefly explain how you arrived at that volume formula?</p>
<p>the original formula is supposed to be volume=1/2<em>length</em>width*height, right?
So, how did you get the h^2?</p>
<p>Thank you a lot for the help.</p>
<p>You want it in terms of h. Since base is 3, height is 1, base and height are in a 3:1 ratio. So base = 3h. </p>
<p>Then area is (10)(3h^2)/2 = 15h^2</p>
<p>gracias (10 char).</p>
<p>so, the derivative of x in the pyt theorem above would be dx/dt instead of 1 correct? and then I factor out the dx/dt...</p>
<p>al bandito, you mean solve for dh/dt correct?</p>