SOLUTION: Find the equation of normal to the curve y= x sqrt(1-2x) at the point (-4,-12).

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Find the equation of normal to the curve y= x sqrt(1-2x) at the point (-4,-12).      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 829382: Find the equation of normal to the curve y= x sqrt(1-2x) at the point (-4,-12).
Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
y = x%2Asqrt%281-2x%29 at the point (-4,-12)

The derivative is the slope of the tangent line.
The slope of the normal is the negative reciprocal of
the slope of the tangent line.

So.
1. We find the derivative.
2. We substitute x=-4 and simplify
3. We take the negative reciprocal of the result of 2
4. We use the point-slope formula with this slope and the given point.

y%22%22=%22%22x%2Asqrt%281-2x%29

matrix%282%2C1%2C%22%22%2Cy%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2Cx%2A%281-2x%29%5E%281%2F2%29%29 


matrix%282%2C1%2C%22%22%2Cdy%2Fdx%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2Cx%2A%281%2F2%29%281-2x%29%5E%28-1%2F2%29%28-2%29%29matrix%282%2C1%2C%22%22%2C%22%22%2B%22%22%29matrix%282%2C1%2C%22%22%2C%281-2x%29%5E%281%2F2%29%281%29%29 


matrix%282%2C1%2C%22%22%2Cdy%2Fdx%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C-x%281-2x%29%5E%28-1%2F2%29%29matrix%282%2C1%2C%22%22%2C%22%22%2B%22%22%29matrix%282%2C1%2C%22%22%2C%281-2x%29%5E%281%2F2%29%29 

We substitute x=-4

matrix%282%2C2%2C%22%22%2C%22%22%2Cdy%2Fdx%2Cat_x=-4%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C-%28-4%29%281-2%28-4%29%29%5E%28-1%2F2%29%29matrix%282%2C1%2C%22%22%2C%22%22%2B%22%22%29matrix%282%2C1%2C%22%22%2C%281-2%28-4%29%29%5E%281%2F2%29%29

matrix%282%2C2%2C%22%22%2C%22%22%2Cdy%2Fdx%2Cat_x=-4%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C4%281%2B8%29%5E%28-1%2F2%29%29matrix%282%2C1%2C%22%22%2C%22%22%2B%22%22%29matrix%282%2C1%2C%22%22%2C%281%2B8%29%5E%281%2F2%29%29

matrix%282%2C2%2C%22%22%2C%22%22%2Cdy%2Fdx%2Cat_x=-4%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C4%289%29%5E%28-1%2F2%29%29matrix%282%2C1%2C%22%22%2C%22%22%2B%22%22%29matrix%282%2C1%2C%22%22%2C%289%29%5E%281%2F2%29%29

matrix%282%2C2%2C%22%22%2C%22%22%2Cdy%2Fdx%2Cat_x=-4%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C4%289%29%5E%28-1%2F2%29%29matrix%282%2C1%2C%22%22%2C%22%22%2B%22%22%29matrix%282%2C1%2C%22%22%2C%289%29%5E%281%2F2%29%29

matrix%282%2C2%2C%22%22%2C%22%22%2Cdy%2Fdx%2Cat_x=-4%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C4%2Fsqrt%289%29%29matrix%282%2C1%2C%22%22%2C%22%22%2B%22%22%29matrix%282%2C1%2C%22%22%2Csqrt%289%29%29

matrix%282%2C2%2C%22%22%2C%22%22%2Cdy%2Fdx%2Cat_x=-4%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C4%2F3%29matrix%282%2C1%2C%22%22%2C%22%22%2B%22%22%29matrix%282%2C1%2C%22%22%2C3%29

matrix%282%2C2%2C%22%22%2C%22%22%2Cdy%2Fdx%2Cat_x=-4%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C4%2F3%29matrix%282%2C1%2C%22%22%2C%22%22%2B%22%22%29matrix%282%2C1%2C%22%22%2C9%2F3%29

matrix%282%2C2%2C%22%22%2C%22%22%2Cdy%2Fdx%2Cat_x=-4%29matrix%282%2C1%2C%22%22%2C%22%22=%22%22%29matrix%282%2C1%2C%22%22%2C13%2F3%29

The slope of the tangent line is 13%2F3, so the slope of the
normal is -3%2F13

So we want the equation of the line through (-4,-12) with slope -3%2F13

y - y1 = m(x - x1)
where (x1,y1) = (-4,-12)

y+-+%28-12%29%22%22=%22%22expr%28-3%2F13%29%28x-%28-4%29%29

y+%2B12%29%22%22=%22%22expr%28-3%2F13%29%28x%2B4%29

Multiply through by 13

13y+156 = -3(x+4)

13y+156 = -3x-12
 
 3x+13y = -168

Here's the graph, it looks like a normal (perpendicular) line.
So it must be correct.

 

Edwin