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- Definition 3.27 Let P(x, y) be a point in R² and 0 be the radian measure of the angle formed by OP (O is the origin) and the positive side of the x-axis. Suppose f is a function of the two variables x and y and U cos 0 î + sin 0 ĵ. The directional derivative of ƒ in the direction of U, denoted by D7#f, is given by f(x + h cos 0, y + h sin 0) – f(x, y) DGf(r, y) = lim h→0 h if this limit exists. Note that the directional derivative is a scalar (a real number). The partial derivatives fr and fy are special cases of the directional derivative. 1 Remark (a) If U = î, then cos 0 = 1 and sin 0 = 0. Thus, f(r + h, y) – f(x,y) D;f(x, y) = lim h-0 h which is the partial derivative of f with respect to x. (6) If U = î, then cos 0 = 0 and sin 0 = 1. Thus, f(x, y + h) – f(r, y) D;f(x, y) = lim h→0 h which is the partial derivative of f with respect to y.arrow_forward5. Find an equation plane to the surface (u, v) = (3 cos(u) sin(v), 2 sin(u) sin(v), cos(u)) at the point (u, v) = (0,7).arrow_forwardc) Verify Stokes's Theorem for F = (x²+y²)i-2xyj takes around the rectangle bounded by the lines x=2, x=-2, y=0 and y=4arrow_forward
- F(1, y.:)= (x², 2x²y*, 5yz*) XZ Given: Find (a) div F (b) curl Farrow_forwardLet F = (3z + 3x²) ¿+ (6y + 4z + 4 sin(y²)) 3+ (3x + 4y + 6e²²) k (a) Find curl F. curl F - (b) What does your answer to part (a) tell you about SF. dr where C' is the circle (x − 15)² + (y – 10)² = 1 in the xy-plane, oriented clockwise? ScF. dr = 0 (c) If C is any closed curve, what can you say about ScF. dr? SoF. dr = 0 (d) Now let C be the half circle - (x − 15)² + (y − 10)² = 1 in the xy-plane with y > 10, traversed from (16, 10) to (14, 10). Find SF. dr by using your result from (c) and considering C plus the line segment connecting the endpoints of C. -0 Lc F. dr =arrow_forwardQ: Solve on white page clean and stepwise: Let f(x, y, z) = sinxyx, g(x, y, z) = x²yz, U = x²yi – y²zj + 2xz²k and V = xi – 2zk. Then verify the following relations 1. div(fVg) = fv²g+Vf•Vg %3D 2. curl(curlU) = grad(divU) – V²Uarrow_forward
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