Question

F(x, y, z) = yzexzi + exzj + xyexzk,

c.r(t) = (t2 + 5)i + (t2 − 1)j + (t2 − 5t)k, 0 ≤ t ≤ 5 (a) find a function f such that f = ∇f.

Answer 1

`\\abla f(x,y,z)=\mathbf F(x,y,z)\iff(\partial f)/(\partial x)\,\mathbf i+(\partial f)/(\partial y)\,\mathbf j+(\partial f)/(\partial z)\,\mathbf k=yze^(xz)\,\mathbf i+e^(xz)\,\mathbf j+xye^(xz)\,\mathbf k`


`(\partial f)/(\partial x)=yze^(xz)`

`\implies f(x,y,z)=\frac{yz}ze^(xz)+g(y,z)=ye^(xz)+g(y,z)`


`(\partial f)/(\partial y)=e^(xz)=e^(xz)+(\partial g)/(\partial y)`

`\implies(\partial g)/(\partial y)=0\implies g(y,z)=h(z)`


`(\partial f)/(\partial z)=xye^(xz)=xye^(xz)+(\mathrm dh)/(\mathrm dz)`

`\implies(\mathrm dh)/(\mathrm dz)=0\implies h(z)=C`


`f(x,y,z)=ye^(xz)+C`


`\mathbf r(t)=(t^2+5)\,\mathbf i+(t^2-1)\,\mathbf j+(t^2-5)\,\mathbf k`

`\implies\mathbf r(0)=5\,\mathbf i-\mathbf j-5\,\mathbf k`

`\implies\mathbf r(5)=30\,\mathbf i+24\,\mathbf j+20\,\mathbf k`

By the gradient theorem, we have for any path
`\mathcal C` originating at the point (5, -1, -5) and terminating at the point (30, 24, 20),


`\displaystyle\int_(\mathcal C)\mathbf F\cdot\mathrm d\mathbf r=f(\mathbf r(5))-f(\mathbf r(0))=f(30,24,20)-f(5,-1,-5)`

`\implies\displaystyle\int_(\mathcal C)\mathbf F\cdot\mathrm d\mathbf r=24e^(600)+e^(-25)`