New questions (1 Viewer)

[currysauce]

A picture 2 metres high is hung on a wall with its bottom edge 6 metres above the eye of the viewer. How far from the wall should the viewer stand for the picture to subtend the largest possible verticle angle with her eye.

2.

Show that cosecx restricted to 0 < x < = pi/2 has an inverse (DONE). Calculate the inverse's derivative (that part)

thanks!

 

[Sober]

Here is my solution. See attached.

I found this way easier.

Let x be the distance from the viewer to the wall and arctan represent inverse tan.

Angle of the picture = y = arctan(8/x) - arctan(6/x)

Let a=arctan(8/x) and b=arctan(6/x)

tan(y) = tan(a-b)

= [ tan(a) - tan(b) ] / [ 1 + tan(a).tan(b) ]

= [ (8/x) - (6/x) ] / [ 1 + (48/x^2) ]

= 2x / (x^2 + 48)

y = arctan( 2x / [x^2 + 48] )

Since arctan is an always increasing function it will be at a maximum when the argument is at a maximum so we need to find the max of:

z = 2x / [x^2 + 48]

z' = ( 2x^2 + 96 - 4x^2 ) / (x^2 + 48)^2

-2x^2 + 96 = 0

x =+ sqrt(48)

There is a maxiumum when x=sqrt(48)

 

[Raginsheep]

Asking year12s to do a uni maths question (even though its "based on ext1 work) is poor effort mate

 

[Raginsheep]

It comes from the calculus problem set for MATH1151 at unsw for actuary and finance/maths students. Its technically not uni work, just something to tie us over until we've covered enough to start on the "real" uni maths problems but still, point stands

 

[currysauce]

thanks guys, its easy now that i think about it...

hey sorry, its been awhile since i have done this stuff... i need a refresher haha. meh

thanks again