Implicit differentiation

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if x^2 + y^2 = xy + 3, find dy/dx where x = 1, and hence find the equations of the two tangents to the curve where x=1.
 
This is calculus, not DE.

Use x=1 in the original equation to find the 2 y-values when x=1.

Differentiate and use the vlaues to find slope at these points.

Use y=mx+b to find the two line equations.
 
Hello, americo74!

Galactus' advice is excellent . . .


If \(\displaystyle x^2\,+\,y^2 \:= \:xy\,+\,3\), (a) find \(\displaystyle \frac{dy}{dx}\) where \(\displaystyle x\,=\,1\).

(b) Find the equations of the two tangents to the curve where \(\displaystyle x\,=\,1\).

When \(\displaystyle x\,=\,1\), the equation is: \(\displaystyle 1\,+\,y^2\:=\:u\,+\,3\)

We have: \(\displaystyle \,y^2\.-\,y\,-\,2\:=\:0\;\;\Rightarrow\;\;(y\,+\,1)(y\,-\,2)\:=\:0\;\;\Rightarrow\;\;y\,=\,-1,\,2\)

The points of tangency are: \(\displaystyle \,(1,\,-1)\) and \(\displaystyle (1,2)\)


Differentiate implicitly: \(\displaystyle \:2x\,+\,2y\left(\frac{dy}{dx}\right) \:=\:x\left(\frac{dy}{dx}\right)\,+\,y\)

. . . . . . . . . . . . . . \(\displaystyle 2y\left(\frac{dy}{dx}\right)\,-\,x\left(\frac{dy}{dx}\right)\:=\:y\,-\,2x\)

. . . . . . . . . . . . . . . . \(\displaystyle (2y\,-\,x)\left(\frac{dy}{dx}\right) \:= \:y\,-\,2x\)

. . . . . . . . . . . . . . . . . . . . . . . \(\displaystyle \frac{dy}{dx} \:= \:\frac{y\,-\,2x}{2y\,-\,x}\)


(a) At \(\displaystyle (1,-1):\;\;\frac{dy}{dx} \:=\:\frac{(-1)\,-\,2(1)}{2(-1)\,-\,1}\:=\:\frac{-3}{-3}\:=\:1\)

. . . At \(\displaystyle (1,2):\;\;\frac{dy}{dx} \:=\:\frac{2-2(1)}{2(2)\,-\,1}\:=\:\frac{0}{3}\:=\:0\)

(b) Now write the equations of the lines with: \(\displaystyle \begin{Bmatrix}(1,-1), & m\,=\,1 \\ (1,2), & m\,=\,0\end{Bmatrix}\)

 
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