Asked by david
Determine equations for the tangents to the points on the curve y = -x^4 + 8x^2 such that the tangent lines are perpendicular to the line x = 1.
I've found the derivative, but not really sure where to go from here.
I've found the derivative, but not really sure where to go from here.
Answers
Answered by
Reiny
The first derivative is the slope of the tangent at a given point
hoping you had
-4x^3 + 16x for the derivative, which, when x = 1 becomes -4+16 or 12
Since the new line is perpendicular to the tangent , it's slope must be -1/12 ,the negative reciprocal.
-4x^3 + 16x = -1/12
4x^3 - 16x - 1/12 = 0
48x^3 - 192x - 1 = 0
perhaps your problem is solving this cubic.
It does not factor , so it is hard to do, and quite messy
You could use something like Wolfram to solve it, .....
It shows 3 real x values, appr. -2, +2, and 0
https://www.wolframalpha.com/input/?i=48x%5E3+-+192x+-+1+%3D+0
I was expecting some nicer numbers, since we are not done yet
pick x = 2.0026, sub this into the original to get y = 15.99989
let's go with (2,16)
so the tangent at the point (2,16) is
y - 16 = (-1/12)(x - 2)
hoping you had
-4x^3 + 16x for the derivative, which, when x = 1 becomes -4+16 or 12
Since the new line is perpendicular to the tangent , it's slope must be -1/12 ,the negative reciprocal.
-4x^3 + 16x = -1/12
4x^3 - 16x - 1/12 = 0
48x^3 - 192x - 1 = 0
perhaps your problem is solving this cubic.
It does not factor , so it is hard to do, and quite messy
You could use something like Wolfram to solve it, .....
It shows 3 real x values, appr. -2, +2, and 0
https://www.wolframalpha.com/input/?i=48x%5E3+-+192x+-+1+%3D+0
I was expecting some nicer numbers, since we are not done yet
pick x = 2.0026, sub this into the original to get y = 15.99989
let's go with (2,16)
so the tangent at the point (2,16) is
y - 16 = (-1/12)(x - 2)
Answered by
david
thanks reiny! my problem was that it was indeed cubic, and I thought i may have perhaps been doing the question wrong. You've helped me on 2 questions today, appreciate it.
Answered by
oobleck
Of course, the maxima are at x = ±√8, so if (2,16) is one point of interest
is (√8-0.828,16) (just to the left of the peak), then the other point is at x = -√8-0.828 = -3.656
is (√8-0.828,16) (just to the left of the peak), then the other point is at x = -√8-0.828 = -3.656
Answered by
oobleck
hmmm. and yet, the cubic works at -2, not -3.656
I'll have to ponder this...
I'll have to ponder this...