SOLUTION: consider the curve y=2logx, where log is the natural logarithm. let α be the tangent to that curve which passes through the origin, let P be the point of contact of α and

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: consider the curve y=2logx, where log is the natural logarithm. let α be the tangent to that curve which passes through the origin, let P be the point of contact of α and      Log On


   

Question 749494: consider the curve y=2logx, where log is the natural logarithm. let α be the tangent to that curve which passes through the origin, let P be the point of contact of α and that curve, and let m be the straight line perpendicular to the tangent α at P. We are to find the equations of the straight lines α and m and the area S of the region bounded by the curve y=2logx, the straight line m, and the x-axis
let t be the x-coordinate of the poin P, then t satisfies log t=(A). Hence the equation of α is
y=%28B%29%2Ax%2Fe
the equation of m is
y=-ex%2F%28C%29+%2B+e%5E2%2F%28D%29+%2B+%28E%29
thus the area S of the region is
S=%28F%29+%2B+%28G%29%2Fe
solve for A,B,C,D,E,F and G
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
The function is y=2%2Aln%28x%29
Its graph crosses the x-axis at the point where y=0
system%28y=2%2Aln%28x%29%2Cy=0%29 --> 2%2Aln%28x%29=0 --> ln%28x%29=0 --> x=1
The x-coordinate of point P is x=t.
The slope of the tangent at point P is the value of the derivative at that point.
y'=2%2Fx, so the slope of the tangent at x=t is 2%2Ft.
Since the line alpha tangent at P passes through the origin, its equation must be
y=%282%2Ft%29x
At point P, with x=t, y=%282%2Ft%29t --> y=2
Since point P is on the graph of y=2%2Aln%28x%29, its y-coordinate is y=2%2Aln%28t%29
So system%28y=2%2Cy=2%2Aln%28t%29%29 --> 2=2%2Aln%28t%29 --> ln%28t%29=1 --> system%28highlight%28A=1%29%2Ct=e%29 and P is (e,2).

Now we can find the equation of alpha:
system%28y=%282%2Ft%29x%2Ct=e%29 --> y=%282%2Fe%29x --> highlight%28B=2%29%0D%0A+%0D%0ASince+the+slope+of+line+%7B%7B%7Balpha is 2%2Fe,
line m perpendicular to alpha must have a slope of -1%2F%282%2Fe%29=-e%2F2.
As m passes through P(e,2) its equation is
y-2=%28-e%2F2%29%28x-e%29 --> y-2=%28-e%2F2%29x%2Be%5E2%2F2 --> y=%28-e%2F2%29x%2Be%5E2%2F2%2B2 --> y=-e%2Ax%2F2%2Be%5E2%2F2%2B2
So highlight%28C=2%29, highlight%28D=2%29, and highlight%28E=2%29
The line m crosses the x-axis at the point where y=0
system%28y=-e%2Ax%2F2%2Be%5E2%2F2%2B2%2Cy=0%29 --> -e%2Ax%2F2%2Be%5E2%2F2%2B2=0 --> -e%2Ax%2Be%5E2%2B4=0 --> e%5E2%2B4=ex --> x=%28e%5E2%2B4%29%2Fe --> x=e%2B4%2Fe

The area S of the region bounded by the curve y=2%2Aln%28x%29, the straight line m, and the x-axis is shown below.
graph%28300%2C300%2C-1%2C6%2C-1%2C6%2C2%2Aln%28x%29%2C-e%2Ax%2F2%2Be%5E2%2F2%2B2%29

S can be can be calculated as the sum of:
the area below y=2%2Aln%28x%29, and above the x-axis, between x=1 and x=e, int%282%2Aln%28x%29%2C+dx%2C+1%2C+e+%29=2%2Aint%28ln%28x%29%2C+dx%2C+1%2C+e+%29
plus the area below m between x=e and x=e%2B4%2Fe, int%28%28-e%2Ax%2F2%2Be%5E2%2F2%2B2%29%2C+dx%2C+e%2C+e%2B4%2Fe+%29

int%28%28-e%2Ax%2F2%2Be%5E2%2F2%2B2%29%2C+dx%2C+e%2C+e%2B4%2Fe+%29 is easier than it seems.
It's just the area of the triangle with vertices (e,0), P(e,2), and (e+e/4,0)
Its base is 4%2Fe; its height is 2, and its area is %281%2F2%29%2A2%2A%284%2Fe%29=4%2Fe.

Since int%28ln%28x%29%2C+dx%29=2%28x%2Aln%28x%29-x%29,


So S=2%2B4%2Fe and highlight%28G=4%29