Let F(x, y) = be a vector field over a region containing C: {(x, y): x^2 + y^2 =

Question

Let F(x, y) = be a vector field over a region containing C: {(x, y): x^2 + y^2 = 1}. Sketch F(0, 0), F(1, 1), F(-1, -1). Determine if F is tangent to C. Let C: {(x, y): y = x^2 + 1, 0 lessthanorequalto x lessthanorequalto 2} oriented from left to Parameterize C. (Write a parametric description of C.) Evaluate: integral _C ds Compute: integral _ - Squareroot 1 - y^2^0 integral _ -1^0 integral _ -1^0 Squareroot 2 - x^2 - y^2 dydzdx Let f(x, y) = 1/Squareroot x^2 + y^2. Find the linear approximation at the point (3, -4).

Explanation / Answer

We first find the equation of the tangent plane at the 'nice' point (3, -4).

f(x, y) = (x^2 + y^2)^(-1/2) ==> f(3, -4) = 0.2
f_x = (-x) / (x^2 + y^2)^(-3/2) ==> f_x (3, 4) = -0.024
f_y = (-y) / (x^2 + y^2)^(-3/2) ==> f_y (3, -4) = 0.032.

Therefore, the equation of the tangent plane is
z - z = f_x(x, y) * (x - x) + f_y (x, y) * (y - y)
==> z - 0.2 = (-0.024)(x - 3) + ( 0.032)(y + 4)
==> z = 0.032y+0.4-0.024x.

For (x, y) sufficiently close to (3 , -4), this forms a good approximation for z = f(x, y).

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