posted by Avik on .
Show that the tangent to the curve 25x^5+5x^4-45x^3-5x^2+2x+6y-24=0 at the point (-1,1) is also a normal at two points of the curve.
125x^4 + 20x^3 - 135x^2 - 10x + 2 + 6dy/dx = 0
125 - 20 - 135 + 10 - 2 + 6dy/dx = 0
6dy/dx = 22, so the slope of the tangent at (-1,1) is 22/6 = 11/3
equation of tangent:
1= (11/3)(-1) + b
b = 1 + 11/3 = 14/3
y = (11/3)x + 14/3 = (11x+14)/3
sub that into the original equation
25x^5+5x^4-45x^3-5x^2 + 24x + 4=0
after dividing out the factor (x+1) we are left with
25x^4 - 20x^3 - 25x^2 +20x+4=0
tried to solve this, even had Wolfram page try it
it gave me x = appr -.9507 and appr -.169872
If we sub those values into the derivative equation we should get -3/11 or -.2727..
I did not get that.
Can't find my error
those two points don't exist
(you might print out my solution and go through it to see if find any errors)
-6y = 25x^5+5x^4-45x^3-5x^2+2x+24
y' = -1/6 (125x^4 + 20x^3 - 135x^2 - 10x + 2)
y'(-1) = 3
normal has slope -1/3
Go to http://rechneronline.de/function-graphs/
-1/6 * (25x^5+5x^4-45x^3-5x^2+2x-24)
and its derivative
as well as
The line is not normal to the curve anywhere. Typo somewhere here?
When plotting the graphs, set
x: -2 to 2
y: -10 to 10
that will show things quite clearly.