Question #184483

11. (Section 7.9 and Chapter 8) Consider the 3–dimensional vector field F defined by F (x, y, z) = ￾ 12x 2 y 2 + 2z 2 + 1, 8x 3 y − 3z, 4xz − 3y − 3  (a) Write down the Jacobian matrix JF (x, y, z). (2) (b) Determine the divergence div F(x, y, z). (2) (c) Determine curl F(x, y, z). (2) (d) Give reasons why F has a potential function. (Refer to the relevant definitions and theorems in the study guide.) (2) (e) Find a potential function of F, using the method of Example 7.9.1. Note, however that that example concerns a 2-dimensional vector field, so you will have to adapt the method to be suitable for a 3-dimensional vector field. Pay special attention to the notation that you use for derivatives of functions of more than one variable.


1
Expert's answer
2021-05-07T09:31:59-0400

Given field,

F(x,y,z)=(12x2y2+2z2+1,8x3y3z,4xz3y3)F(x,y,z)=(12x^2y^2+2z^2+1,8x^3y-3z,4xz-3y-3)


or F=(12x2y2+2z2+1)i^+(8x3y3z)j^+(4xz3y3)k^\vec{F}=(12x^2y^2+2z^2+1)\hat{i}+(8x^3y-3z)\hat{j}+(4xz-3y-3)\hat{k}


(1) Jacobian Matrix J(F(x,y,z)=[dfxdxdfxdydfxdzdfydxdfydydfydzdfzdxdfzdydfzdz]J(F(x,y,z)=\begin{bmatrix} \dfrac{df_x}{dx} & \dfrac{df_x}{dy}& \dfrac{df_x}{dz} \\\\ \dfrac{df_y}{dx}& \dfrac{df_y}{dy}& \dfrac{df_y}{dz}\\\\ \dfrac{df_z}{dx} & \dfrac{df_z}{dy}& \dfrac{df_z}{dz} \end{bmatrix}



=[24xy224x2y4z24xy28x334z30]\begin{bmatrix} 24xy^2& 24x^2y& 4z \\\\ 24xy^2 & 8x^3& -3\\\\ 4z& -3&0 \end{bmatrix}


(2) divF(x,y,z)div F(x, y, z)


=(ddxi^+ddyj^+ddzk^).((12x2y2+2z2+1)i^+(8x3y3z)j^+(4xz3y3)k^)= (\dfrac{d}{dx}\hat{i}+\dfrac{d}{dy}\hat{j}+\dfrac{d}{dz}\hat{k}).((12x^2y^2+2z^2+1)\hat{i}+(8x^3y-3z)\hat{j}+(4xz-3y-3)\hat{k})


=24xy2+8x3+4x=24xy^2+8x^3+4x


(3) curlF(x,y,z)=[i^j^k^ddxddyddz12x2y2+2z2+18x3y3z4xz3y3]curl F(x,y,z)=\begin{bmatrix} \hat{i} & \hat{j} &\hat{k}\\\\ \dfrac{d}{dx} &\dfrac{d}{dy}&\dfrac{d}{dz} \\\\ 12x^2y^2+2z^2+1& 8x^3y-3z& 4xz-3y-3 \end{bmatrix}


=i^(3(3))j^(4z4z)+k^(24x2y24x2y)=0=\hat{i}(-3-(-3))-\hat{j}(4z-4z)+\hat{k}(24x^2y-24x^2y)\\ =0


(4) F is a potential function because, Whenever There exist a field, There exist a potential associated

with it. We can calculate the potential function F by the integraing the given vector field.


(5) As we know-

Potential function is given by-

Fv=F.drF_v=-\int F.dr


=((12x2y2+2z2+1)i^+(8x3y3z)j^+(4xz3y3)k^)(dxi^+dyj^+dzz^)=\int ((12x^2y^2+2z^2+1)\hat{i}+(8x^3y-3z)\hat{j}+(4xz-3y-3)\hat{k})(dx\hat{i}+dy\hat{j}+dz\hat{z})


=(4x3y+2z2x+x+4x3y23yz+2x2z3xy3x)=(4x^3y+2z^2x+x+4x^3y^2-3yz+2x^2z-3xy-3x)


=8x3y2+2x2z+2xz23xy3yz2x=8x^3y^2+2x^2z+2xz^2-3xy-3yz-2x


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS