Question

1.12 marks Suppose the measurements of ore lode in a deposit of gold are described by the func- L(x,y,z)-(10 32-y - 3y +0.4ry), where z is the vertical coordinate, with z = 0 at the surface and z < 0 being below ground. The tion (z+3)2 distances are measured in tens of metres (a) Draw contours of the function at a depth of z-3. Use Matlab, Octave or similar. You might like to include the range [x,y] E (-5,5) (b) The edge of the ore body is where the recording shows L 0. Find the equation for this boundary if z =-3. Write this equation in the explicit form y (c) A miner is at the point (2,1,-3). How is the ore lode changing at this point (2,1,-3) if they dig vertically downward? (d) In what direction should they "dig" to head toward the highest concentration of gold? (e) If they dig in the direction b (3,-4,1) what is the rate of increase in gold concentration? (f) Are the results in (d), (e) consistent with your contour plot? Show the vectors on the plot.

Answer 1

1. (a) At z = -3, the equation becomes,

At z = -3, the equation becomes, L(x, y, -3) = 10 - 3x^2 - y^2 - 3y + 0.4xy Plotting the contour we get, (b) When L = 0, we can write, implies 10 - 3x^2 - y^2 - 3y + 0.4xy = 0 implies 3x^2 + y62 + 3y - 0.4xy = 10 implies 15x^2 + 5y^2 + 15y = 2xy Solving for y we get, (Solved in MATLAB using solve function applied to above equation) implies (y - 0.2x + 1.5)^2 = (-3x^2 - 0.6x + 2.22)^2 implies y = 0.2x - 1.5 plusminus squareroot -3x^2 - 0.6x + 2.22 (c) The function is, L(x, y, z) = e^-(z + 3)^2(10 - 3x^2 - y^2 - 3y + 0.4xy) Digging downwards implies that we have to take the negative partial derivative of z and substitute (2,1,-3) to get the rate of change, This gives, partial differential L/partial differential z = e^(-z + 3)^2 (-2(z + 3))(10 - 3x62 - y^2 - 3y + 0.4xy) Substituting (2,1,-3) we get, partial differential L/partial differential z = e^-(-3 + 3)^2 (-2(-3

Plotting the contour we get,

(b) When L = 0, we can write,

$\Rightarrow 10 - 3x^2 - y^2 - 3y + 0.4xy = 0$

$\Rightarrow 3x^2 + y^2 + 3y - 0.4xy = 10$

$\Rightarrow 15x^2 + 5y^2 + 15y = 2xy$

Solving for y we get, (Solved in MATLAB using solve function applied to above equation)

$\Rightarrow y = 0.2x-1.5 \pm \sqrt{-3x^2 - 0.6x + 2.22}$

(c) The function is,

Digging downwards implies that we have to take the negative partial derivative of z and substitute (2,1,-3) to get the rate of change, This gives,

Substituting (2,1,-3) we get,

Hence there is no change, L is constant.

(d) For the direction for highest concentration of gold, we have to take the gradient of the above scalar function. We get the gradient as,

$\triangledown L = \frac{\partial L}{\partial x} \, \widehat{i} + \frac{\partial L}{\partial y} \, \widehat{j} + \frac{\partial L}{\partial z} \, \widehat{k}$

$\Rightarrow \frac{\partial L}{\partial x} = -e^{-(z + 3)^2}(6x - 0.4y)$

$\Rightarrow \frac{\partial L}{\partial y} = -e^{-(z + 3)^2}(2y - 04y + 3)$

$\Rightarrow \frac{\partial L}{\partial z} = -e^{-(z + 3)^2}(2z + 6)(3x^2 - 0.4xy + y^2 + 3y - 10)$

Hence the gradient is given by,

$\Rightarrow (-e^{-(z + 3)^2}(6x - 0.4y))\widehat{i} + (-e^{-(z + 3)^2}(2y - 04y + 3))\widehat{j} + (-e^{-(z + 3)^2}(2z + 6)(3x^2 - 0.4xy + y^2 + 3y - 10))\widehat{k}$

Substituting the point (2,1,-3) we get the direction as,

$\Rightarrow -11.6 \,\widehat{i} -4.2 \, \widehat{j}$