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
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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
the equation of m is
thus the area S of the region is
solve for A,B,C,D,E,F and G
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
The function is
Its graph crosses the x-axis at the point where
--> --> -->
The x-coordinate of point P is .
The slope of the tangent at point P is the value of the derivative at that point.
y'=, so the slope of the tangent at is .
Since the line tangent at P passes through the origin, its equation must be
At point P, with , -->
Since point P is on the graph of , its y-coordinate is
So --> --> --> and P is (e,2).
Now we can find the equation of :
--> --> is ,
line perpendicular to must have a slope of .
As passes through P(e,2) its equation is
--> --> -->
So , , and
The line crosses the x-axis at the point where
--> --> --> --> -->
The area of the region bounded by the curve , the straight line , and the x-axis is shown below.
can be can be calculated as the sum of:
the area below , and above the x-axis, between and ,
plus the area below between and ,
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 ; its height is , and its area is .
Since ,
So and
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