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?

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!

thank you very much, I understand my mistake!

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)

To determine the coordinates of point Q that maximizes the angle of vision of an observer, situated at the origin, following the movement of point Q along the curve, you can indeed use derivatives. However, in this case, you need to find the derivative of the angle with respect to the position of point Q. Let's go through the process step by step:

1. Start by finding the equation of the curve represented by y = f(x) = (2x - 10)^(1/2).

2. Determine the slope of the curve at any point (x, y) on the curve. To do this, take the derivative of f(x) with respect to x, which gives:

f'(x) = (1/2)(2x - 10)^(-1/2) * 2 = (2x - 10)^(-1/2)

3. Now, consider a line joining point Q (x, y) on the curve to the origin (0, 0). The observer at the origin sees an angle, let's call it theta (θ), which is given by the formula tan(theta) = y/x.

4. To find the angle of vision, we need to express theta in terms of x. So, divide y by x:

tan(theta) = y/x = f(x)/x = [(2x - 10)^(1/2)]/x

5. To maximize the angle of vision, we need to maximize tan(theta). Therefore, we need to find the maximum value of [(2x - 10)^(1/2)]/x.

6. Take the derivative of [(2x - 10)^(1/2)]/x with respect to x:

d/dx [(2x - 10)^(1/2)]/x = [(2x - 10)^(-1/2)] - [(2x - 10)^(1/2)]/x^2

7. Set this derivative equal to zero and solve for x to find critical points. This will help us identify the x-coordinate of point Q that maximizes the angle of vision.

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

8. Solve the equation to find the x-coordinate(s) of the critical point(s).

9. Once you have the x-coordinate(s) of the critical point(s), plug the values into the equation y = (2x - 10)^(1/2) to find the corresponding y-coordinate(s) of point Q.

10. The coordinates of point Q that maximize the angle of vision are determined by the values obtained in step 9.

Remember that maximizing the angle of vision involves finding the maximum value of [(2x - 10)^(1/2)]/x, which corresponds to the critical point(s) identified in step 8.