1. ## slope of tangent

Use "implicit differentiation" to find the slope of tangent when x= sqrt 3 for x^2 + y^2 = 4

i did

dx^2/dx + dy^2/dx = d4/dx

dx^2/dx = 2x
dy^2/dx = 2yy'
d4/dx = 0

2x + 2yy' = 0
y' = - x/y

and then what do i have to do to find the slope of the tangent ?

2. Originally Posted by haebinpark
Use "implicit differentiation" to find the slope of tangent when x= sqrt 3 for x^2 + y^2 = 4

i did

dx^2/dx + dy^2/dx = d4/dx

dx^2/dx = 2x
dy^2/dx = 2yy'
d4/dx = 0

2x + 2yy' = 0
y' = x/y

and then what do i have to do to find the slope of the tangent ?
All you have to do is solve the original equation $\displaystyle x^2+y^2=4$ for $\displaystyle y$ and the substitute into the equation for $\displaystyle y'$.

$\displaystyle 3+y^2=4$

$\displaystyle y=1$

$\displaystyle y'=\frac{x}{y}=\sqrt{3}$

3. isn't y should be plus minus 1 ?
how can you tell if it is +1 ?

and i got y' = - x / y ?

hm... did i do something wrong?

4. Originally Posted by haebinpark
isn't y should be plus minus 1 ?
how can you tell if it is +1 ?

and i got y' = - x / y ?

hm... did i do something wrong?
Actually, I you're right.

$\displaystyle 3+y^2=4$
$\displaystyle y=\pm1$
$\displaystyle y'=\frac{x}{y}=\pm\sqrt{3}$

Notice that if you graph the circle, that there are two values of $\displaystyle y$ that correspond to the point $\displaystyle x=\sqrt{3}$. Both of these points $\displaystyle (\sqrt{3},\pm 1)$ will have tangents that have opposite slopes.

5. thank you !!!
helped a lot !!

so slope of tangent will be +/- sqrt 3. right?