I believe this is for a calculus class. I assume you have been paying at least a tiny bit of attention in class :)
a) Finding the RoC of x when y = 50
Take the derivative of [tex] \sqrt{100^2+y^2} [/tex]
[tex]=(100^2+y^2)^\frac{1}{2}[/tex]
By Chain Rule, we can drop the 1/2 exponent to the front. The exponent then becomes 1/2 - 1 = -1/2. Then multiply the derivative of the inside function.
[tex]\frac{1}{2} (100^2 + y^2)^ \frac{-1}{2} \cdot(2y)[/tex]
Simplifying we get,
[tex]\frac{d}{dy} (100^2+y^2)^ \frac{1}{2} = \frac{y}{ \sqrt{10000+y^2} } [/tex]
Plug y = 50 into your derivative.
[tex] \frac{50}{ \sqrt{12500}}[/tex]
around 0.447
b) Find the RoC of the area of triangle ABC when y = 50
Find the derivative of [tex] \frac{1}{2} \cdot100\cdot y = 50y[/tex]
[tex] \frac{d}{dy}[50y]=50 [/tex]
The change of area is always 50, regardless of the y value.
c) Find the RoC of theta when y = 50
Let O = theta
sin O = y/100
[tex]O = sin^{-1} ( \frac{y}{100})[/tex]
Take the derivative of O using Chain Rule
The derivative of [tex]sin^{-1}(g(x)) = \frac{1}{ \sqrt{1- g^{2}(x) } } \cdot \frac{d}{dx}(g(x)) [/tex]
Thus,
[tex] \frac{d}{dy} [sin^{-1} ( \frac{y}{100})]= \frac{1}{ \sqrt{1- \frac{y^2}{10000} } } \cdot \frac{1}{100} [/tex]
Plug y = 50 back in.
[tex] \frac{1}{ \sqrt{1- \frac{2500}{10000} } } \cdot \frac{1}{100} [/tex]
which is around 0.0115
