Question

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 1

Answer:

Answer is $f(1.01)=0.91$

Step-by-step explanation:

Since (1,1) lies on the curve it must satisfy it hence

$x^{2}+sin(xy)+3y^{2}=C\\1+sin(1)+3=C\\\therefore C=4.017$

Now tangent line approximation of $f(x)$ is given by

$f(x)=f(x_o)+f'(x_o)(x-x_o)$

Applying values we get

$f(1.01)=f(1)+f'(1)(1.01-1)$

Now differentiating $x^{2}+sin(xy)+3y^{2}=C$ we get

$2x+cos(xy)\times (y+x\frac{dy}{dx})+6y=0$

Solving for $\frac{dy}{dx}$ we get

$\frac{dy}{dx}=\frac{1}{x}(\frac{-6y-2x}{cos(xy)}-y)$

Applying values we get

$\frac{dy}{dx}=-9.00$

Using all the values we have obtained we get

$f(1.01)=1-9.00\times (1.01-1)f(1.01)=0.91$