Final answer:
To find the equation of the tangent plane to the graph of f(x,y)=ln(11x^2-8y^2) at the point (1,1), we need to find the partial derivatives of f(x,y) and substitute the given point into these derivatives. Using the partial derivatives, we can determine the equation of the tangent plane as z-(ln(3/100))=(22/3)(x-1)-(16/3)(y-1).
Step-by-step explanation:
To find the equation of the tangent plane to the graph of f(x,y)=ln(11x2-8y2) at the point (1,1), we need to find the values of fx, fy, and f(1,1). Taking the partial derivatives, we get fx=22/(11x2-8y2) and fy=(-16y)/(11x2-8y2). Substituting (1,1) into these equations, we get fx=22/3 and fy=-16/3.
So, the equation of the tangent plane is given by (z-f(1,1))=fx(x-1)+fy(y-1), which simplifies to z-(ln(3/100))=(22/3)(x-1)-(16/3)(y-1).Using the partial derivatives, we can determine the equation of the tangent plane as z-(ln(3/100))=(22/3)(x-1)-(16/3)(y-1).