SOLUTION: Find the equation of the tangent line to the given curves. 1. y=(7x-6)^1/3 that is perpendicular to the line 12x+7y+2=0

Algebra ->  Linear-equations -> SOLUTION: Find the equation of the tangent line to the given curves. 1. y=(7x-6)^1/3 that is perpendicular to the line 12x+7y+2=0      Log On


   

Question 1136445: Find the equation of the tangent line to the given curves.
1. y=(7x-6)^1/3 that is perpendicular to the line 12x+7y+2=0
Answer by greenestamps(13203) About Me  (Show Source):
You can put this solution on YOUR website!


The slope of the given line is -12/7; the slope of a line perpendicular to it has a slope of 7/12.

The given curve is y+=+%287x-6%29%5E%281%2F3%29

The derivative is dy%2Fdx+=+%281%2F3%29%287%29%287x-6%29%5E%28-2%2F3%29+=+7%2F%283%287x-6%29%5E%282%2F3%29%29

We need to find the value of x for which the derivative (slope) is 7/12:

7%2F%283%287x-6%29%5E%282%2F3%29%29+=+7%2F12
3%287x-6%29%5E%282%2F3%29+=+12
%287x-6%29%5E%282%2F3%29+=+4
7x-6+=+4%5E%283%2F2%29+=+8
7x+=+14
x+=+2

We need to determine the y value of the curve at x=2:

y+=+%287%282%29-6%29%5E%281%2F3%29+=+8%5E%281%2F3%29+=+2

We want the equation of the line with slope 7/12 passing through (2,2):

2+=+%287%2F12%292%2Bb
2+=+7%2F6%2Bb
b+=+5%2F6

The equation of our tangent line is

y+=+%287%2F12%29x%2B5%2F6

A graph....

graph%28400%2C400%2C-2%2C6%2C-4%2C4%2C%287x-6%29%5E%281%2F3%29%2C%287%2F12%29x%2B5%2F6%29