SOLUTION: The second derivative of a curve is y'' = 12x^2 − 8. The tangent to this curve at (4, 192) is perpendicular to the line x + 224y − 448 = 0. Find the equation of the cur
Algebra ->
Test
-> SOLUTION: The second derivative of a curve is y'' = 12x^2 − 8. The tangent to this curve at (4, 192) is perpendicular to the line x + 224y − 448 = 0. Find the equation of the cur
Log On
Question 1075331: The second derivative of a curve is y'' = 12x^2 − 8. The tangent to this curve at (4, 192) is perpendicular to the line x + 224y − 448 = 0. Find the equation of the curve. Found 2 solutions by Boreal, Edwin McCravy:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! y"=12x^2-8
y'=4x^3-8x+C1
y=x^4-4x^2+C1x+C2
When x=4 y=256-64+4C1+C2
4C1+C2=0
The slope of the line tangent to this is perpendicular to 224y=-x+448; y=-x/224+2, and it is 224, the negative reciprocal
point-slope y-y1=m(x-x1) m slope (x1,y1) point
y-192=224(x-4)
y=224x-704
The slope of 224=4x^3-8x+C1, where x=4
224=256-32+C1, so C1=0, C2=0
The curve is x^4-4x^2, or x^2(x+2)(x-2)
The slope of the line x + 224y − 448 = 0 is found by taking the derivative
implicitly 1 + 224y' - 0 = 0
224y' = -1
y' = <--- slope
Since the tangent to the curve at (4,192) has slope which is perpendicular
to that, it has slope which is its negative reciprocal or +224.
So we set y' = 224, and x = 4 in
Substitution C=0 in
Since the curve passes through the point (4, 192) we substitute
x=4 and y=192
So to find the equation we substitute C=0 in
Edwin
