Friday

January 30, 2015

January 30, 2015

Posted by **Cristy** on Wednesday, August 1, 2012 at 10:00pm.

- Calculus -
**Steve**, Thursday, August 2, 2012 at 5:30amthe line touching the graph at x=a has slope -2/a^3

The line through (a,1/a^2) with slope -2/a^3 is

y = -2/a^3 (x-a) + 1/a^2

This line crosses the axes at (0,3/a^2) and (3a/2,0)

So, the area of the triangle formed by the line and the axes is

A(a) = 1/2 * 3/a^2 * 3a/2 = 9/(4a)

Now we have us a problem. dA/da is never 0. The area is infinite at a=0, and goes to zero at a = oo

**Answer this Question**

**Related Questions**

calculus - A line in the first quadrant is tangent to the graph of y=1/x^2. How ...

math - A tangent line is drawn to the hyerbola xy=c at a point P. 1) show that ...

math repost!! - please help, i procrastinated and now this is due tomorrow!! A ...

Calculus - Let f be the function given by f(x)=(x^3)/4 - (x^2)/3 - x/2 + 3cosx. ...

Calculus - I am doing the AP calculus review, these are the questions I have no ...

calculus - Find the area of the largest rectangle with one corner at the origin...

Calculus - If a tangent line is drawn to the parabola y = 3 - x^2 at any point ...

calculus - If a tangent line is drawn to the parabola y = 3 - x^2 at any point ...

Calc. - Find the area of the region bounded by the parabola y=x^2, the tangent ...

calculus - Let A be the area if the region in the first quadrant under the graph...