Calculus
posted by Jim on .
With y=f(x) (2x10)^1/2
having a virtual line joined at point Q from the origin (0,0), determine the coordinates of point Q that maximises the angle of vision of an observer, situated at the origin, following the movement of point Q along the curve.
Note that tan (theta symbol) = y/x
Will I just have to find both derivatives and from there find the zero for the maximum?

Did you make a sketch of y = (2x10)^(1/2) ?
Let Q be any point on the curve
then Q can be labelled (x, (2x10^1/2 )
Join OQ
slope of oQ = (2x10)^1/2 / x
let the angle made by OQ with the xaxis be Ø
then
tanØ = y/x = (2x10)^1/2 / x
sec^2 Ø dØ/dx = (x(1/2)(2x10)^(1/2)(2)  (2x10)^(1/2) (1) )/x^2 using the quotient rule
for a max of Ø , dØ/dx = 0
so (x(1/2)(2x10)^(1/2)(2)  (2x10)^(1/2) (1) )/x^2 = 0
(x(1/2)(2x10)^(1/2)(2)  (2x10)^(1/2) (1) ) = 0 , we can ignore the denominator.
x/(2x10)^(1/2)  (2x10)^1/2 = 0
x/(2x10)^1/2 = (2x10)^1/2
x = 2x10 , after crossmultiplying
x = 10
so Q is (10, 10^1/2) or (10, √10)
check:
at x=10, tanØ = √10/10 = .31622 , Ø = 17.548°
at x = 10.1 , tanØ = √10.2/10.1 = .3162 Ø = 17.547 °  a bit smaller
at x = 9.9 , tanØ = √9.8/9.9 = .3162116, Ø = 17.547  again a bit smaller than at x=10
my answer looks good! 
thank you very much, I understand my mistake!

So, we want to maximize
t = arctan(sqrt(2x10)/x)
t' = (x10)/[(sqrt(2x10)/x)^2 + 1)*x^2 * sqrt(2x10)]
All that junk in the bottom is never zero when x>5, so we just have to find where x10 = 0
well, duh: x=10
So, Q = (10,√10)