find tangent vector to the graph
\[2x^2 + y^2 = 9 \] at (2,1)
@wio
Differentiate
\[ 4x+2yy'=0 \]
\[ 4(2)+2(1)y'=0 \]
\[ y'=-4 \]
ok wait y prime = -2x/y
so how do we change it to tangent vector
Just gonna be \(\langle 1,-4\rangle\) I believe.
i mean how, i know how u got y prime how about x prime
I differentiated with respect to \(x\), so \(D_x x=1\)
gotcha thanks and <1,-4> is tangent vector right?
You could use the chain rule and get: \[ 4xx'+2yy′=0 \]
ofcourse it is tangent vector as it is the derivative.
\[ 8x'+2y'=0 \]Think of it like a dot product: \[ \langle 8,2 \rangle \cdot \langle x',y' \rangle = 0 \]
Remember to get an orthogonal vector to \(\langle a,b \rangle\) you just use \(\langle -b,a \rangle\)
makes sense but i need tangent vector so by this new dot product method how do we approach <1,-4>
Well \[ \langle 8,2\rangle \cdot \langle 1,-4\rangle =0 \]
<8,2> * <x',y'> = 0 ----> <1,-4> ?????
Yeah
Or use \[ 2y'=-8x' \]Let \(x'=1\) and then:\[ 2y'=-8\implies y'=-4 \]
how do u know that x prime is 1
Why not?
idk why cant it be -1 or 0
Well it can't be 0 because the 0 vector has no direction
You can use -1
-1 will give us y prime as 4
which is not right
Sure it is, they're the same direction
ohk wait i m confused can u explain me <8,2> * <x',y'> = 0 from this method how to get x prime = 1 and y' = -4
You didn't specify a magnitude. If you did then it would matter, ore than just direction.
Just let \(x'=1\) or any other value you want it to be and solve for \(y'\). There are infinite solutions.
actually my real question is to find directional derivative in direction of tangent vector to this graph
so does it matter if i use any of <1,-4> or <-1, 4>
No
It doesn't matter. Those both point in the same proportional direction.
ok so i will do the dot product of <1,-4> and gradient thanks
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