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 $\displaystyle g(x,t)=x^2\cos(2t)$
taking some partials we get
$\displaystyle g_x(x,t)=2x\cos(2t) \iff g_x(2,\pi/2)=2(2)\cos(\pi)=-4$
$\displaystyle g_t(x,t)=-2x^2\sin(2t) \iff g_t(2,\pi/2)=-2(2)^2\sin(\pi)=0$
$\displaystyle g(2,\pi/2)=(2)^2\cos(\pi)=-4$
so finally our approximation (a,b) is
$\displaystyle L(x,t)=g(a,b)+g_x(a,b)(x-a)+g_t(a,b)(t-b)$
pluggin in our values gives
$\displaystyle L(x,t)=-4-4(x-2)=-4x+4$