local linearization

**dexza666** · April 13th 2008, 04:52 PM

hi how do you find the local linearization of the function g(x,t) = x2cos(2t) at the point (2,P/2)

i have no idea

**TheEmptySet** · April 13th 2008, 05:28 PM

Originally Posted by dexza666

hi how do you find the local linearization of the function g(x,t) = x2cos(2t) at the point (2,P/2)

i have no idea

They are asking you to find the tangent plane at the point

I guess you mean this function $g(x,t)=x^2\cos(2t)$

taking some partials we get

$g_x(x,t)=2x\cos(2t) \iff g_x(2,\pi/2)=2(2)\cos(\pi)=-4$

$g_t(x,t)=-2x^2\sin(2t) \iff g_t(2,\pi/2)=-2(2)^2\sin(\pi)=0$

$g(2,\pi/2)=(2)^2\cos(\pi)=-4$

so finally our approximation (a,b) is

$L(x,t)=g(a,b)+g_x(a,b)(x-a)+g_t(a,b)(t-b)$

pluggin in our values gives

$L(x,t)=-4-4(x-2)=-4x+4$

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