SOLUTION: A curve passes through he point P(0,3.5) and is such that dy/dx=2-x. The normal to the curve at P meets the curve again at Q. Find the coordianates of Q.

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Question 1044364: A curve passes through he point P(0,3.5) and is such that dy/dx=2-x. The normal to the curve at P meets the curve again at Q.
Find the coordianates of Q.
Found 2 solutions by ikleyn, robertb:
Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website!
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A curve passes through he point P(0,3.5) and is such that dy/dx=2-x. The normal to the curve at P meets the curve again at Q.
Find the coordinates of Q.
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Since you are Calculus, I will talk to you as to a Calculus person.

1. Since dy/dx = 2-x, y = - + , where C is a constant. Find C from the condition that y(0) = 3.5. (Use the info about the point P). 2. The normal to the curve at P is the straight line y - x = k, where "k" is a constant. (Why ?) Find the constant "k" from the condition that the normal passes through the point P = (0,3.5). 3. Now solve the system y = - + , y - x = k to find the second intersection point Q.

Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website!
By solving the differential equation, we easily get
,
as per the boundary condition.
Now, to get the slope of the tangent line at the point (0,7/2), simply substitute x = 0 into the original DE ,
from which we get .
===> The slope of the normal line to the curve at point P is -1/2 (i.e., the negative reciprocal of 2).
===> The equation of the normal line to the curve at point P is
, or <===THE NORMAL LINE
To find the other point Q on the curve where the normal line meets the curve again, solve
,
===> ===> 0 = x(5-x) ===> x = 0, 5.
When x = 0, y = 7/2, which is just the first point the normal meets the curve.
When x = 5, y = 1, which is the point (5,1), the second point where the normal meets the curve.