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

Algebra ->  Linear-equations -> 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      Log On


   



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) About Me  (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 y=x%5E2 at P(h,h2).
The derivative is dy%2Fdx=2x, so the slope at P(h,h2) is 2h.
Since it goes through (h,h2), its equation is:

y-h%5E2=2h%28x-h%29

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

0-h%5E2=2h%28x-h%29
-h%5E2=2hx-2h%5E2%29
Divide through by h
-h=2x-2h
h=2x
expr%281%2F2%29h=x

So the coordinates of Q are Q%28+expr%281%2F2%29h%2C0%29.



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 expr%281%2F2%29x.  So let the variable
q = the x-coordinate of Q.  q=expr%281%2F2%29x. We want to know %28dq%29%2F%28dt%29 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

ds%2Fdt+=+2

%28ds%2Fdt%29%5E2=%28dx%2Fdt%29%5E2%2B%28dy%2Fdt%29%5E2

2%5E2=%28dx%2Fdt%29%5E2%2B%282x%2Aexpr%28dx%2Fdt%29%29%5E2

4=%28dx%2Fdt%29%5E2%2B4x%5E2%2A%28dx%2Fdt%29%5E2
   
4=%28dx%2Fdt%29%5E2%281%2B4x%5E2%29

4%5E%22%22%2F%281%2B4x%5E2%29=%28dx%2Fdt%29%5E2

dx%2Fdt=sqrt%284%5E%22%22%2F%281%2B4x%5E2%29%29

dx%2Fdt=2sqrt%281%5E%22%22%2F%281%2B4x%5E2%29%29

q=expr%281%2F2%29x
dq%2Fdt+=+expr%281%2F2%29%28dx%2Fdt%29
dq%2Fdt+=+expr%281%2F2%29%282sqrt%281%5E%22%22%2F%281%2B4x%5E2%29%29%29
dq%2Fdt+=+sqrt%281%5E%22%22%2F%281%2B4x%5E2%29%29

Evaluating that when x=2,

%22%22=%22%221%2Fsqrt%2817%29 units per second.

Edwin


Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
*
If x+=+sqrt%28y%29, then 1+=+%28dy%2Fdx%29%2F%282%2Asqrt%28y%29%29, or dy%2Fdx+=+2sqrt%28y%29+=+2x
If we let point P be (x%5B0%5D, y%5B0%5D), then the tangent line is
y+-+y%5B0%5D+=+%28dy%28x%5B0%5D%29%2Fdx%29%2A%28x-x%5B0%5D%29, whose x-intercept is obtained by letting y = 0:
-y%5B0%5D+=+2sqrt%28y%5B0%5D%29%28x-x%5B0%5D%29 ===> -sqrt%28y%5B0%5D%29%2F2+=+x-x%5B0%5D <===> -x%5B0%5D%2F2+=+x-x%5B0%5D ===> x%5Bi%5D+=+x%5B0%5D%2F2.
===> With respect to time, dx%5Bi%5D%2Fdt+=+%281%2F2%29%28dx%5B0%5D%2Fdt%29, or 2%28dx%5Bi%5D%2Fdt%29+=+dx%5B0%5D%2Fdt.
Now the arc length "s" from x = 0 to x = x%5B0%5D is given by
.

===> ds%2Fdt+=+sqrt%281%2B+4x%5B0%5D%5E2%29%2A%28dx%5B0%5D%2Fdt%29%29, by a direct application of the fundamental theorem of calculus.

===> ds%2Fdt+=+sqrt%281%2B+4x%5B0%5D%5E2%29%2A2%2A%28dx%5Bi%5D%2Fdt%29%29.

Since at the instant when P=(2,4) ds%2Fdt+=+2, we get
2=+sqrt%281%2B+4x%5B0%5D%5E2%29%2A2%2A%28dx%5Bi%5D%2Fdt%29%29, so that dx%5Bi%5D%2Fdt+=+1%2F+sqrt%281%2B+4%2A2%5E2%29+=+highlight%281%2Fsqrt%2817%29%29 unit per second,

the rate of change of the x-intercept Q.