For a college football field with the dimension shown, the angle for kicking a field goal from a ( horizontal) distance of x feet from the goal post is given by∅(x)=tan^(-1)⁡〖(29.25/x〗)-〖tan^(-1) (〗⁡〖10.75/x〗) .Show that f(t)=t/(a^3+t^2 ) is increasing for a>t

Could you have a clearer equation?

No

To show that f(t) = t/(a^3 + t^2) is increasing for a > t, we need to demonstrate that the derivative of f(t) with respect to t is positive for all values of t where a > t.

Let's start by calculating the derivative of f(t) using the quotient rule:

f'(t) = [ (a^3 + t^2)(1) - t(2t) ] / (a^3 + t^2)^2
= (a^3 + t^2 - 2t^2) / (a^3 + t^2)^2
= (a^3 - t^2) / (a^3 + t^2)^2

Now, we need to determine the conditions under which f'(t) > 0. Since we are given that a > t, let's evaluate f'(t) for a particular value of t, such that t < a:

f'(t) = (a^3 - t^2) / (a^3 + t^2)^2

Since a > t, both a^3 and t^2 are positive. Therefore, a^3 - t^2 is also positive. Additionally, (a^3 + t^2)^2 is always positive since it's a square.

Thus, the sign of f'(t) is determined by the positive numerator only. Since the numerator (a^3 - t^2) is positive for a > t, it follows that f'(t) > 0 for a > t.

As a result, we have shown that f(t) = t/(a^3 + t^2) is increasing for a > t, as its derivative f'(t) is positive.