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Mathematics 22 Online
OpenStudy (anonymous):

find tangent vector to the graph

OpenStudy (anonymous):

\[2x^2 + y^2 = 9 \] at (2,1)

OpenStudy (goformit100):

@wio

OpenStudy (anonymous):

Differentiate

OpenStudy (anonymous):

\[ 4x+2yy'=0 \]

OpenStudy (anonymous):

\[ 4(2)+2(1)y'=0 \]

OpenStudy (anonymous):

\[ y'=-4 \]

OpenStudy (anonymous):

ok wait y prime = -2x/y

OpenStudy (anonymous):

so how do we change it to tangent vector

OpenStudy (anonymous):

Just gonna be \(\langle 1,-4\rangle\) I believe.

OpenStudy (anonymous):

i mean how, i know how u got y prime how about x prime

OpenStudy (anonymous):

I differentiated with respect to \(x\), so \(D_x x=1\)

OpenStudy (anonymous):

gotcha thanks and <1,-4> is tangent vector right?

OpenStudy (anonymous):

You could use the chain rule and get: \[ 4xx'+2yy′=0 \]

OpenStudy (anonymous):

ofcourse it is tangent vector as it is the derivative.

OpenStudy (anonymous):

\[ 8x'+2y'=0 \]Think of it like a dot product: \[ \langle 8,2 \rangle \cdot \langle x',y' \rangle = 0 \]

OpenStudy (anonymous):

Remember to get an orthogonal vector to \(\langle a,b \rangle\) you just use \(\langle -b,a \rangle\)

OpenStudy (anonymous):

makes sense but i need tangent vector so by this new dot product method how do we approach <1,-4>

OpenStudy (anonymous):

Well \[ \langle 8,2\rangle \cdot \langle 1,-4\rangle =0 \]

OpenStudy (anonymous):

<8,2> * <x',y'> = 0 ----> <1,-4> ?????

OpenStudy (anonymous):

Yeah

OpenStudy (anonymous):

Or use \[ 2y'=-8x' \]Let \(x'=1\) and then:\[ 2y'=-8\implies y'=-4 \]

OpenStudy (anonymous):

how do u know that x prime is 1

OpenStudy (anonymous):

Why not?

OpenStudy (anonymous):

idk why cant it be -1 or 0

OpenStudy (anonymous):

Well it can't be 0 because the 0 vector has no direction

OpenStudy (anonymous):

You can use -1

OpenStudy (anonymous):

-1 will give us y prime as 4

OpenStudy (anonymous):

which is not right

OpenStudy (anonymous):

Sure it is, they're the same direction

OpenStudy (anonymous):

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

OpenStudy (anonymous):

You didn't specify a magnitude. If you did then it would matter, ore than just direction.

OpenStudy (anonymous):

Just let \(x'=1\) or any other value you want it to be and solve for \(y'\). There are infinite solutions.

OpenStudy (anonymous):

actually my real question is to find directional derivative in direction of tangent vector to this graph

OpenStudy (anonymous):

so does it matter if i use any of <1,-4> or <-1, 4>

OpenStudy (anonymous):

No

OpenStudy (anonymous):

It doesn't matter. Those both point in the same proportional direction.

OpenStudy (anonymous):

ok so i will do the dot product of <1,-4> and gradient thanks

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