## nickymarden Group Title Determine the tangent line to he curve y^2(2-x)=x^3 at the point P(1,1). 2 years ago 2 years ago

1. myko Group Title

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

2. nickymarden Group Title

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

3. myininaya Group Title

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

4. nickymarden Group Title

yeah, i did that :)

5. myininaya Group Title

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

6. myininaya Group Title

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

7. myininaya Group Title

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

8. nickymarden Group Title

thank you :))

9. robtobey Group Title

$(2-x) y^2=x^3$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)$-\frac{\left(3\ 1^2+1^2\right)}{2 (1-2) 1}=2$