Probem20+Julie,+Kevin+W.

Kevin Wu:
 * 1) 20. Find, correct to two decimal places, the coordinates of the point on the curve y=tanx that is closest to the point (1,1).

d=[(x-1)^2+(y-1)^2]^(1/2) d=[(x-1)^2+(tanx-1)^2]^(1/2) d^2 = f(x)=(x-1)^2+(tanx-1)^2 f ' (x) = 2(x-1) + 2(tanx-1)(sec(x)^2) f ' (x) = 2x-2 + 2tanxsec(x)^2 - 2sec(x)^2 f ' (x) = x-1+tanxsec(x)^2-sec(x)^2 f ' (x) = x-1 + tanx(tan^2x +1) - (tan^2x + 1) f ' (x) = x-1 + tan^3(x) + tanx - tan^2(x)

plug into calc.

x=.84

we need to find what value x= when f '(x)=0, however I had trouble doing this on my calculator because I could not find one number to equal 0 for this function. Try this again...I plugged this function into my Y1 and looked for where it crossed the x-axis. I did not get x = 1!!! (ms sweeney) Based on my calculator I found that when x=1 y=1.557 which is very close to

If you plug x-1 + tan^3(x) + tanx - tan^2(x) into Y1 and looked for where it crossed the x-axis, you should get x=.542 y=0.