Determine the tangent line to he curve y^2(2-x)=x^3 at the point P(1,1).

Question

anonymous · Answer

put it like this:
$$y=\pm \sqrt{x ^{3}/(2-x)}$$
abd find the derivative

anonymous · Answer

i did $$f(1) =1$$ $$(f(x))^2.(2-x)=x^3$$ and derived it

myininaya · Answer

$$y^2(2-x)=x^3$$
$$(y^2)'(2-x)+y^2(2-x)'=(x^3)' \text{ product rule }$$
$$2yy'(2-x)+y^2(0-1)=3x^2$$

anonymous · Answer

yeah, i did that :)

myininaya · Answer

$$2yy'(2-x)+y^2(-1)=3x^2$$
$$2yy'(2-x)-y^2=3x^2$$

myininaya · Answer

$$\text{ add } y^2 \text{ on both sides }$$
$$\text{ divide both sides by } 2y(2-x) $$

myininaya · Answer

The objective is to solve for y'
Following these steps will do that
y' is the slope by the way

anonymous · Answer

thank you :))

anonymous · Answer

$$(2-x) y^2=x^3 $$Calculate the total derivative of the above$$2 (2-x) y dy-y^2 dx=3x^2 dx $$Solve for dy$$dy=-\frac{\left(3 x^2+y^2\right) dx}{2 (x-2) y} $$then divide both sides by dx$$\frac{dy}{ dx}=-\frac{\left(3 x^2+y^2\right)}{2 (x-2) y} $$Evaluate the RHS of the above at point (1,1)$$-\frac{\left(3\ 1^2+1^2\right)}{2 (1-2) 1}=2 $$