SOLUTION: A line tangent to the curve x = y^(1/2) at the point P intersects the x axis at the point Q. If P travels up the curve at a rate of 2 units per second, how fast is the point Q

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Question 1183948: A line tangent to the curve x = y^(1/2) at the point P intersects the x axis at
the point Q.
If P travels up the curve at a rate of 2 units per second, how fast is the point
Q traveling when P passes through the point (2,4)?

Found 2 solutions by Edwin McCravy, robertb:
Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!


Suppose the line is tangent to the curve at P(h,h2). 

Then its slope is the derivative of  at P(h,h2).
The derivative is , so the slope at P(h,h2) is 2h.
Since it goes through (h,h2), its equation is:



Its x-intercept P is the x-value when y=0, so



Divide through by h




So the coordinates of Q are .



Now let's switch the letter h to x.  [It would have been too confusing
if we had started out with x because the equation of a line uses x).



We want to know how fast Q is moving.  Since Q is on the x-axis, Q moves the
same speed as its own x-coordinate .  So let the variable
q = the x-coordinate of Q.  . We want to know  at P(2,4) which is when x=2.

The point P moves up the curve at the rate of 2 units per second, so








   













Evaluating that when x=2,

 units per second.

Edwin


Answer by robertb(5830)   (Show Source): You can put this solution on YOUR website!
*
If , then , or
If we let point P be (, ), then the tangent line is
, whose x-intercept is obtained by letting y = 0:
===> <===> ===> .
===> With respect to time, , or .
Now the arc length "s" from x = 0 to x = is given by
.

===> , by a direct application of the fundamental theorem of calculus.

===> .

Since at the instant when P=(2,4) , we get
, so that unit per second,

the rate of change of the x-intercept Q.

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