Question

Hi, I need the full worked solution/explanations for ALL parts of this question please. Please do not skip steps and underline all final answers. The final answers to all parts are provided below the question. Neat handwriting is greatly appreciated. Thank you! :)

Question 4 Consider the contour curve given by (а) Identify a unit vector perpendicular to the curve at the point (-1, 1) (4 marks) (b) Consider the scalar field fx, y, )y-y+xyz Identify the maximum value of the derivative of this scalar field f at the point (1, 2, , and a unit vector that specifies the direction in which this maximum value occurs (5 marks) (c) Find the divergence of the vector field F(x, y, ) xyi-y?j+xyz k. (3 marks) Final Answers: 1 4. (a) (-2i3) V13 (b) (61- 6k) yL-2yzi+xy (c)

Answer 1

a) Given f(x,y) = x²y³-1 = 0

The gradient of f(x,y) at point (x,y) is a vector normal to the contour curve at this point.

Given point is (-1,1)

The gradient is obtained as follows

∇f(x,y)= (fx,fy) =

  $= (\frac{\partial }{\partial x} f(x,y) i + \frac{\partial }{\partial y} f(x,y)j )$

= (2xy³i+ 3x²y²j)

= (2*-1*1³ i +  3*-1²*1²j)

= (-2i+3j)

and corresponding unit vector will be

= (-2i+3j) /(√2²+3²)

= (-2i+3j) /(√13)

c)

vector field F=Pi+Qj+Rk = xy²i-y²z²j+xyzk

here P = xy², Q = -y²z², R= xyz

divergence is defined as

div F= (∂P/∂x) i+ (∂Q/∂y) j+ (∂R/∂z) z

div F= y²i - 2yz² j + xy k