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.
-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/
and plot
-1/6 * (25x^5+5x^4-45x^3-5x^2+2x-24)
and its derivative
as well as
3x+4
The line is not normal to the curve anywhere. Typo somewhere here?
whewww!
125x^4 + 20x^3 - 135x^2 - 10x + 2 + 6dy/dx = 0
at (-1,1)
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+2x+6y-24=0
25x^5+5x^4-45x^3-5x^2+2x+6(11x+14)/3-24=0
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
http://www.wolframalpha.com/input/?i=25x%5E4+-+20x%5E3+-+25x%5E2+%2B20x%2B4%3D0
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
OR,
those two points don't exist
(you might print out my solution and go through it to see if find any errors)
When plotting the graphs, set
x: -2 to 2
y: -10 to 10
that will show things quite clearly.
To determine whether the tangent to the curve at (-1,1) is also a normal at two points on the curve, we need to find the slope of the tangent at (-1,1) and then check if that slope corresponds to two other points on the curve.
Let's start by finding the slope of the tangent at (-1,1). To do this, we'll take the derivative of the given equation implicitly with respect to x.
The given equation is:
25x^5 + 5x^4 - 45x^3 - 5x^2 + 2x + 6y - 24 = 0
Differentiating both sides with respect to x:
125x^4 + 20x^3 - 135x^2 - 10x + 2 + 6y' = 0
Next, substitute the coordinates of the point (-1,1) into the equation:
125(-1)^4 + 20(-1)^3 - 135(-1)^2 - 10(-1) + 2 + 6y' = 0
Simplifying the equation:
125 - 20 - 135 + 10 + 2 + 6y' = 0
-18 + 6y' = 0
Solving for y':
6y' = 18
y' = 18/6
y' = 3
The slope of the tangent at (-1,1) is 3. Therefore, we have the equation of the tangent line: y = 3(x + 1) + 1
y = 3x + 4
To find the points at which the tangent line is also the normal, we need to solve the equation of the curve and the equation of the tangent line simultaneously. The resulting points will be the solutions to this system of equations.
Substitute the equation of the tangent line into the equation of the curve:
25x^5 + 5x^4 - 45x^3 - 5x^2 + 2x + 6(3x + 4) - 24 = 0
Simplifying the equation:
25x^5 + 5x^4 - 45x^3 - 5x^2 + 2x + 18x + 24 - 24 = 0
25x^5 + 5x^4 - 45x^3 - 5x^2 + 20x = 0
Now, we need to solve this polynomial equation to find the x-coordinates of the points of intersection. This can be done using numerical methods or software.
Once you find the x-coordinates of the points of intersection, substitute them back into the equation of the tangent line (y = 3x + 4) to find the corresponding y-coordinates.
The two points where the tangent line is also a normal to the curve should satisfy both the equation of the curve and the equation of the tangent line.
Note: Solving the polynomial equation can be quite complex, so it is recommended to use software or numerical methods for obtaining accurate results.