Posted by **Jim** on Tuesday, November 15, 2011 at 5:06pm.

With y=f(x) (2x-10)^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?

- Calculus -
**Reiny**, Tuesday, November 15, 2011 at 6:35pm
Did you make a sketch of y = (2x-10)^(1/2) ?

Let Q be any point on the curve

then Q can be labelled (x, (2x-10^1/2 )

Join OQ

slope of oQ = (2x-10)^1/2 / x

let the angle made by OQ with the x-axis be Ø

then

tanØ = y/x = (2x-10)^1/2 / x

sec^2 Ø dØ/dx = (x(1/2)(2x-10)^(-1/2)(2) - (2x-10)^(1/2) (1) )/x^2 using the quotient rule

for a max of Ø , dØ/dx = 0

so (x(1/2)(2x-10)^(-1/2)(2) - (2x-10)^(1/2) (1) )/x^2 = 0

(x(1/2)(2x-10)^(-1/2)(2) - (2x-10)^(1/2) (1) ) = 0 , we can ignore the denominator.

x/(2x-10)^(-1/2) - (2x-10)^1/2 = 0

x/(2x-10)^1/2 = (2x-10)^1/2

x = 2x-10 , after cross-multiplying

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!

- Calculus -
**Jim**, Tuesday, November 15, 2011 at 6:36pm
thank you very much, I understand my mistake!

- Calculus -
**Steve**, Tuesday, November 15, 2011 at 6:46pm
So, we want to maximize

t = arctan(sqrt(2x-10)/x)

t' = -(x-10)/[(sqrt(2x-10)/x)^2 + 1)*x^2 * sqrt(2x-10)]

All that junk in the bottom is never zero when x>5, so we just have to find where x-10 = 0

well, duh: x=10

So, Q = (10,√10)

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