Question

Please help me with the below question.

Answer 1

a)
`F(x,y,z)` is conservative if it is the gradient field for some scalar function
`f(x,y,z)`. This would require

`(\partial f)/(\partial x) = e^(y + 2z)`

`(\partial f)/(\partial y) = x \, e^(y + 2z)`

`(\partial f)/(\partial z) = ax \, e^(y+2z)`

Integrating both sides of the first equation with respect to
`x` yields

`f(x,y,z) = x \, e^(y + 2z) + g(y, z)`

Differentiate with respect to
`y` :

`(\partial f)/(\partial y) = x \, e^(y + 2z) = x \, e^(y + 2z) + (\partial g)/(\partial y) \implies (\partial g)/(\partial y) = 0 \implies g(y, z) = h(z)`

Differentiate with respect to
`z` :

`(\partial f)/(\partial z) = ax \, e^(y + 2z) = 2x \, e^(y + 2z) + (dh)/(dz)`

We want
`h(z)` to be independent of
`x` and
`y`; we can make them both disappear by picking
`\boxed{a=2}`.

b) This is the so-called triple product, which has the property

`(7\,\hat\jmath - 4\,\hat k) \cdot \bigg((-2\,\hat\imath+3\,\hat k) * (\hat\imath + 2\,\hat\jmath-\hat k)\bigg) = \det \begin{bmatrix} 0 & 7 & -4 \\ -2 & 0 & 3 \\ 1 & 2 & -1 \end{bmatrix}`

Computing the determinant is easy with a cofactor expansion along the first column:

`\det \begin{bmatrix} 0 & 7 & -4 \\ -2 & 0 & 3 \\ 1 & 2 & -1 \end{bmatrix} = 2 \det \begin{bmatrix} 7 & -4 \\ 2 & -1 \end{bmatrix} + \det \begin{bmatrix} 7 & -4 \\ 0 & 3 \end{bmatrix} \\\\ = 2 + 21 = \boxed{23}`

c) Let

`z = f(x, y) = \frac{2x^2 + 2xy - 1}3`

Compute the partial derivatives and evaluate them at
`x=y=1` :

`(\partial f)/(\partial x) = \frac{4x + 2y}3 \implies f_x(1,1) = 2`

`(\partial f)/(\partial y) = \frac{2x}3 \implies f_y(1,1) = \frac23`

Then the tangent plane to
`f(x,y)` at (1, 1, 1) has equation

`z - 1 = 2 (x - 1) + \frac23 (y - 1) \implies \boxed{6x + 2y - 3z = 5}`

d) In polar coordinates,
`R` is the set

`R = \left\{ (r, \theta) ~:~ 0 \le r \le 3 \text{ and } 0 \le \theta \le 2\pi \right\}`

Then the integral evaluates to

`\displaystyle \iint_R (dA)/(\pi√(x^2+y^2)) = \frac1\pi \int_0^(2\pi) \int_0^3 (r\,dr\,d\theta)/(√(r^2)) \\\\ = \frac1\pi \left(\int_0^(2\pi)d\theta\right) \left(\int_0^3dr\right) = 2*3 = \boxed{6}`

e) By the chain rule,

`(\partial w)/(\partial t) = (\partial w)/(\partial x) (\partial x)/(\partial t) + (\partial w)/(\partial y) (\partial y)/(\partial t)`

Eliminating the parameter, we find

`(x + y^2) - (3x + y) = t - t \implies x = \frac{y^2 - y}2`

so that
`x=1` when
`y=2`.

Compute derivatives:

`(\partial w)/(\partial x) = 14x^(13)`

`(\partial w)/(\partial y) = 1`

`\begin{cases} x + y^2 = t \\ 3x + y = t \end{cases} \implies \begin{cases}(\partial x)/(\partial t) + 2y (\partial y)/(\partial t) = 1 \\\\ 3 (\partial x)/(\partial t) + (\partial y)/(\partial t) = 1 \end{cases}`

`\left((\partial x)/(\partial t) + 2y(\partial y)/(\partial t)\right) - 2y \left(3(\partial x)/(\partial t) + (\partial y)/(\partial t)\right) = 1 - 2y \implies (\partial x)/(\partial t) = (2y-1)/(6y-1)`

`3\left((\partial x)/(\partial t) + 2y (\partial y)/(\partial t)\right) - \left(3(\partial x)/(\partial t) + (\partial y)/(\partial t)\right) = 3 - 1 \implies (\partial y)/(\partial t) = \frac2{6y-1}`

Then at the point (1, 1), the derivative we want is

`(\partial w)/(\partial t) = 14 *\frac3{11} + \frac2{11} = (44)/(11) = \boxed{4}`