a) [tex]F(x,y,z)[/tex] is conservative if it is the gradient field for some scalar function [tex]f(x,y,z)[/tex]. This would require
[tex]\dfrac{\partial f}{\partial x} = e^{y + 2z}[/tex]
[tex]\dfrac{\partial f}{\partial y} = x \, e^{y + 2z}[/tex]
[tex]\dfrac{\partial f}{\partial z} = ax \, e^{y+2z}[/tex]
Integrating both sides of the first equation with respect to [tex]x[/tex] yields
[tex]f(x,y,z) = x \, e^{y + 2z} + g(y, z)[/tex]
Differentiate with respect to [tex]y[/tex] :
[tex]\dfrac{\partial f}{\partial y} = x \, e^{y + 2z} = x \, e^{y + 2z} + \dfrac{\partial g}{\partial y} \implies \dfrac{\partial g}{\partial y} = 0 \implies g(y, z) = h(z)[/tex]
Differentiate with respect to [tex]z[/tex] :
[tex]\dfrac{\partial f}{\partial z} = ax \, e^{y + 2z} = 2x \, e^{y + 2z} + \dfrac{dh}{dz}[/tex]
We want [tex]h(z)[/tex] to be independent of [tex]x[/tex] and [tex]y[/tex]; we can make them both disappear by picking [tex]\boxed{a=2}[/tex].
b) This is the so-called triple product, which has the property
[tex](7\,\hat\jmath - 4\,\hat k) \cdot \bigg((-2\,\hat\imath+3\,\hat k) \times (\hat\imath + 2\,\hat\jmath-\hat k)\bigg) = \det \begin{bmatrix} 0 & 7 & -4 \\ -2 & 0 & 3 \\ 1 & 2 & -1 \end{bmatrix}[/tex]
Computing the determinant is easy with a cofactor expansion along the first column:
[tex]\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}[/tex]
c) Let
[tex]z = f(x, y) = \dfrac{2x^2 + 2xy - 1}3[/tex]
Compute the partial derivatives and evaluate them at [tex]x=y=1[/tex] :
[tex]\dfrac{\partial f}{\partial x} = \dfrac{4x + 2y}3 \implies f_x(1,1) = 2[/tex]
[tex]\dfrac{\partial f}{\partial y} = \dfrac{2x}3 \implies f_y(1,1) = \dfrac23[/tex]
Then the tangent plane to [tex]f(x,y)[/tex] at (1, 1, 1) has equation
[tex]z - 1 = 2 (x - 1) + \dfrac23 (y - 1) \implies \boxed{6x + 2y - 3z = 5}[/tex]
d) In polar coordinates, [tex]R[/tex] is the set
[tex]R = \left\{ (r, \theta) ~:~ 0 \le r \le 3 \text{ and } 0 \le \theta \le 2\pi \right\}[/tex]
Then the integral evaluates to
[tex]\displaystyle \iint_R \frac{dA}{\pi\sqrt{x^2+y^2}} = \frac1\pi \int_0^{2\pi} \int_0^3 \frac{r\,dr\,d\theta}{\sqrt{r^2}} \\\\ = \frac1\pi \left(\int_0^{2\pi}d\theta\right) \left(\int_0^3dr\right) = 2\times3 = \boxed{6}[/tex]
e) By the chain rule,
[tex]\dfrac{\partial w}{\partial t} = \dfrac{\partial w}{\partial x} \dfrac{\partial x}{\partial t} + \dfrac{\partial w}{\partial y} \dfrac{\partial y}{\partial t}[/tex]
Eliminating the parameter, we find
[tex](x + y^2) - (3x + y) = t - t \implies x = \dfrac{y^2 - y}2[/tex]
so that [tex]x=1[/tex] when [tex]y=2[/tex].
Compute derivatives:
[tex]\dfrac{\partial w}{\partial x} = 14x^{13}[/tex]
[tex]\dfrac{\partial w}{\partial y} = 1[/tex]
[tex]\begin{cases} x + y^2 = t \\ 3x + y = t \end{cases} \implies \begin{cases}\dfrac{\partial x}{\partial t} + 2y \dfrac{\partial y}{\partial t} = 1 \\\\ 3 \dfrac{\partial x}{\partial t} + \dfrac{\partial y}{\partial t} = 1 \end{cases}[/tex]
[tex]\left(\dfrac{\partial x}{\partial t} + 2y\dfrac{\partial y}{\partial t}\right) - 2y \left(3\dfrac{\partial x}{\partial t} + \dfrac{\partial y}{\partial t}\right) = 1 - 2y \implies \dfrac{\partial x}{\partial t} = \dfrac{2y-1}{6y-1}[/tex]
[tex]3\left(\dfrac{\partial x}{\partial t} + 2y \dfrac{\partial y}{\partial t}\right) - \left(3\dfrac{\partial x}{\partial t} + \dfrac{\partial y}{\partial t}\right) = 3 - 1 \implies \dfrac{\partial y}{\partial t} = \dfrac2{6y-1}[/tex]
Then at the point (1, 1), the derivative we want is
[tex]\dfrac{\partial w}{\partial t} = 14 \times\dfrac3{11} + \dfrac2{11} = \dfrac{44}{11} = \boxed{4}[/tex]
