Question

What steps can I take to get to the answer? What are the formulas that I'll need.

Answer 1

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
`√(100^2+y^2)`

`=(100^2+y^2)^(1)/(2)`
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.

`(1)/(2) (100^2 + y^2)^ (-1)/(2) \cdot(2y)`
Simplifying we get,

`(d)/(dy) (100^2+y^2)^ (1)/(2) = (y)/( √(10000+y^2) )`
Plug y = 50 into your derivative.

`(50)/( √(12500))`
around 0.447

b) Find the RoC of the area of triangle ABC when y = 50
Find the derivative of
`(1)/(2) \cdot100\cdot y = 50y`

`(d)/(dy)[50y]=50`
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

`O = sin^(-1) ( (y)/(100))`
Take the derivative of O using Chain Rule
The derivative of
`sin^(-1)(g(x)) = \frac{1}{ \sqrt{1- g^(2)(x) } } \cdot (d)/(dx)(g(x))`
Thus,

`(d)/(dy) [sin^(-1) ( (y)/(100))]= \frac{1}{ \sqrt{1- (y^2)/(10000) } } \cdot (1)/(100)`
Plug y = 50 back in.

`\frac{1}{ \sqrt{1- (2500)/(10000) } } \cdot (1)/(100)`
which is around 0.0115