Consider the curve given by x^2+ sin(xy)+3y^2 = C , where C is a constant. The point (1, 1) lies on this curve. Use the tangent line approximation to approximate the y-coordinate when x = 1.01. A. 0.996 B. 1 C. 1.004 D. Cannot be determined E. 1.388

Answer:Answer is [tex]f(1.01)=0.91[/tex]Step-by-step explanation:Since (1,1) lies on the curve it must satisfy it hence [tex]x^{2}+sin(xy)+3y^{2}=C\\1+sin(1)+3=C\\\therefore C=4.017[/tex]Now tangent line approximation of [tex]f(x)[/tex] is given by[tex]f(x)=f(x_o)+f'(x_o)(x-x_o)[/tex]Applying values we get[tex]f(1.01)=f(1)+f'(1)(1.01-1)[/tex]Now differentiating [tex]x^{2}+sin(xy)+3y^{2}=C[/tex] we get[tex]2x+cos(xy)\times (y+x\frac{dy}{dx})+6y=0[/tex]Solving for [tex]\frac{dy}{dx}[/tex] we get [tex]\frac{dy}{dx}=\frac{1}{x}(\frac{-6y-2x}{cos(xy)}-y)[/tex]Applying values we get[tex]\frac{dy}{dx}=-9.00[/tex]Using all the values we have obtained we get [tex]f(1.01)=1-9.00\times (1.01-1)f(1.01)=0.91[/tex]