Partial Derivatives of $f(x,y,z)$ : For Question I part i, we need to find the partial derivatives $f_x$ and $f_y$ of $f(x,y,z) = x^2 \sin(xyz)$ . To find $f_x$ , differentiate with respect to $x$ while treating $y$ and $z$ as constants. $f_x = \frac{d}{dx} [x^2 \sin(xyz)] = 2x \sin(xyz) + x^2 \cos(xyz) \cdot yz$
Implicit Differentiation: Now, to find $f_y$ , differentiate with respect to $y$ while treating $x$ and $z$ as constants. $\newline$ $f_y = \frac{d}{dy} [x^2 \sin(xyz)] = x^2 \cos(xyz) \cdot xz$
Double Integral Evaluation: For Question I part ii, we need to find the partial derivatives $f_x$ and $f_y$ of $f(x,y) = 3\ln(x^3 + y^3)$ . To find $f_x$ , differentiate with respect to $x$ while treating $y$ as a constant. $f_x = \frac{d}{dx} [3\ln(x^3 + y^3)] = 3 \cdot \frac{(3x^2)}{(x^3 + y^3)}$
Line Integral Evaluation: To find $f_y$ , differentiate with respect to $y$ while treating $x$ as a constant. $\newline$ $f_y = \frac{d}{dy} [3\ln(x^3 + y^3)] = 3 \cdot \frac{(3y^2)}{(x^3 + y^3)}$
Homogeneous Differential Equation Solution: For Question $2$ , we need to use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ from the equation $3x^2yz^2 - x \sin y + 5xy = 3$ . Differentiating with respect to x, we get: $6xyz^2 + 3x^2y\frac{dz}{dx}z - \sin y + 5y = 0$ Now, solve for $\frac{dz}{dx}$ . $\frac{dz}{dx} = \frac{\sin y - 5y - 6xyz^2}{3x^2yz}$
Homogeneous Differential Equation Solution: For Question $2$ , we need to use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ from the equation $3x^2yz^2 - x \sin y + 5xy = 3$ . Differentiating with respect to $x$ , we get: $6xyz^2 + 3x^2y\frac{dz}{dx}z - \sin y + 5y = 0$ Now, solve for $\frac{dz}{dx}$ . $\frac{dz}{dx} = \frac{\sin y - 5y - 6xyz^2}{3x^2yz}$ Differentiating with respect to $y$ , we get: $3x^2z^2 + 3x^2z\frac{dz}{dy}z - x \cos y + 5x = 0$ Now, solve for $\frac{dz}{dy}$ . $\frac{\partial z}{\partial y}$ $0$
Homogeneous Differential Equation Solution: For Question $2$ , we need to use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ from the equation $3x^2yz^2 - x \sin y + 5xy = 3$ . Differentiating with respect to x, we get: $6xyz^2 + 3x^2y\frac{dz}{dx}z - \sin y + 5y = 0$ Now, solve for $\frac{dz}{dx}$ . $\frac{dz}{dx} = \frac{\sin y - 5y - 6xyz^2}{3x^2yz}$ Differentiating with respect to y, we get: $3x^2z^2 + 3x^2z\frac{dz}{dy}z - x \cos y + 5x = 0$ Now, solve for $\frac{dz}{dy}$ . $\frac{dz}{dy} = \frac{x \cos y - 5x - 3x^2z^2}{3x^2z^2}$ For Question $3$ , we need to apply a double integral to evaluate the function $f(x,y) = x^2y^2 + x$ over the region bounded by $\frac{\partial z}{\partial y}$ $0$ , $\frac{\partial z}{\partial y}$ $1$ , $\frac{\partial z}{\partial y}$ $2$ , and $\frac{\partial z}{\partial y}$ $3$ . Set up the double integral as: $\frac{\partial z}{\partial y}$ $4$ First, integrate with respect to y: $\frac{\partial z}{\partial y}$ $5$
Homogeneous Differential Equation Solution: For Question $2$ , we need to use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ from the equation $3x^2yz^2 - x \sin y + 5xy = 3$ . Differentiating with respect to $x$ , we get: $6xyz^2 + 3x^2y\frac{dz}{dx}z - \sin y + 5y = 0$ Now, solve for $\frac{dz}{dx}$ . $\frac{dz}{dx} = \frac{\sin y - 5y - 6xyz^2}{3x^2yz}$ Differentiating with respect to $y$ , we get: $3x^2z^2 + 3x^2z\frac{dz}{dy}z - x \cos y + 5x = 0$ Now, solve for $\frac{dz}{dy}$ . $\frac{\partial z}{\partial y}$ $0$ For Question $3$ , we need to apply a double integral to evaluate the function $\frac{\partial z}{\partial y}$ $1$ over the region bounded by $\frac{\partial z}{\partial y}$ $2$ , $\frac{\partial z}{\partial y}$ $3$ , $\frac{\partial z}{\partial y}$ $4$ , and $\frac{\partial z}{\partial y}$ $5$ . Set up the double integral as: $\frac{\partial z}{\partial y}$ $6$ First, integrate with respect to $y$ : $\frac{\partial z}{\partial y}$ $8$ Now, evaluate the inner integral: $\frac{\partial z}{\partial y}$ $9$ Simplify the expression: $3x^2yz^2 - x \sin y + 5xy = 3$ $0$
Homogeneous Differential Equation Solution: For Question $2$ , we need to use implicit differentiation to find $(\frac{\partial z}{\partial x})$ and $(\frac{\partial z}{\partial y})$ from the equation $3x^2yz^2 - x \sin y + 5xy = 3$ . Differentiating with respect to $x$ , we get: $6xyz^2 + 3x^2y(\frac{dz}{dx})z - \sin y + 5y = 0$ Now, solve for $(\frac{dz}{dx})$ . $(\frac{dz}{dx}) = \frac{\sin y - 5y - 6xyz^2}{3x^2yz}$ Differentiating with respect to $y$ , we get: $3x^2z^2 + 3x^2z(\frac{dz}{dy})z - x \cos y + 5x = 0$ Now, solve for $(\frac{dz}{dy})$ . $(\frac{\partial z}{\partial y})$ $0$ For Question $3$ , we need to apply a double integral to evaluate the function $(\frac{\partial z}{\partial y})$ $1$ over the region bounded by $(\frac{\partial z}{\partial y})$ $2$ , $(\frac{\partial z}{\partial y})$ $3$ , $(\frac{\partial z}{\partial y})$ $4$ , and $(\frac{\partial z}{\partial y})$ $5$ . Set up the double integral as: $(\frac{\partial z}{\partial y})$ $6$ First, integrate with respect to $y$ : $(\frac{\partial z}{\partial y})$ $8$ Now, evaluate the inner integral: $(\frac{\partial z}{\partial y})$ $9$ Simplify the expression: $3x^2yz^2 - x \sin y + 5xy = 3$ $0$ Finally, integrate with respect to $x$ : $3x^2yz^2 - x \sin y + 5xy = 3$ $2$ Evaluate the definite integral: $3x^2yz^2 - x \sin y + 5xy = 3$ $3$
Homogeneous Differential Equation Solution: For Question $2$ , we need to use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ from the equation $3x^2yz^2 - x \sin y + 5xy = 3$ . Differentiating with respect to $x$ , we get: $6xyz^2 + 3x^2y\frac{dz}{dx}z - \sin y + 5y = 0$ Now, solve for $\frac{dz}{dx}$ . $\frac{dz}{dx} = \frac{\sin y - 5y - 6xyz^2}{3x^2yz}$ Differentiating with respect to $y$ , we get: $3x^2z^2 + 3x^2z\frac{dz}{dy}z - x \cos y + 5x = 0$ Now, solve for $\frac{dz}{dy}$ . $\frac{\partial z}{\partial y}$ $0$ For Question $3$ , we need to apply a double integral to evaluate the function $\frac{\partial z}{\partial y}$ $1$ over the region bounded by $\frac{\partial z}{\partial y}$ $2$ , $\frac{\partial z}{\partial y}$ $3$ , $\frac{\partial z}{\partial y}$ $4$ , and $\frac{\partial z}{\partial y}$ $5$ . Set up the double integral as: $\frac{\partial z}{\partial y}$ $6$ First, integrate with respect to $y$ : $\frac{\partial z}{\partial y}$ $8$ Now, evaluate the inner integral: $\frac{\partial z}{\partial y}$ $9$ Simplify the expression: $3x^2yz^2 - x \sin y + 5xy = 3$ $0$ Finally, integrate with respect to $x$ : $3x^2yz^2 - x \sin y + 5xy = 3$ $2$ Evaluate the definite integral: $3x^2yz^2 - x \sin y + 5xy = 3$ $3$ For Question $4$ , we need to apply a line integral to evaluate $3x^2yz^2 - x \sin y + 5xy = 3$ $4$ over the line segment $3x^2yz^2 - x \sin y + 5xy = 3$ $5$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $6$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $7$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $8$ . First, find $3x^2yz^2 - x \sin y + 5xy = 3$ $9$ , which is the differential arc length: $x$ $0$ $x$ $1$ $x$ $2$
Homogeneous Differential Equation Solution: For Question $2$ , we need to use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ from the equation $3x^2yz^2 - x \sin y + 5xy = 3$ . Differentiating with respect to $x$ , we get: $6xyz^2 + 3x^2y\frac{dz}{dx}z - \sin y + 5y = 0$ Now, solve for $\frac{dz}{dx}$ . $\frac{dz}{dx} = \frac{\sin y - 5y - 6xyz^2}{3x^2yz}$ Differentiating with respect to $y$ , we get: $3x^2z^2 + 3x^2z\frac{dz}{dy}z - x \cos y + 5x = 0$ Now, solve for $\frac{dz}{dy}$ . $\frac{\partial z}{\partial y}$ $0$ For Question $3$ , we need to apply a double integral to evaluate the function $\frac{\partial z}{\partial y}$ $1$ over the region bounded by $\frac{\partial z}{\partial y}$ $2$ , $\frac{\partial z}{\partial y}$ $3$ , $\frac{\partial z}{\partial y}$ $4$ , and $\frac{\partial z}{\partial y}$ $5$ . Set up the double integral as: $\frac{\partial z}{\partial y}$ $6$ First, integrate with respect to $y$ : $\frac{\partial z}{\partial y}$ $8$ Now, evaluate the inner integral: $\frac{\partial z}{\partial y}$ $9$ Simplify the expression: $3x^2yz^2 - x \sin y + 5xy = 3$ $0$ Finally, integrate with respect to $x$ : $3x^2yz^2 - x \sin y + 5xy = 3$ $2$ Evaluate the definite integral: $3x^2yz^2 - x \sin y + 5xy = 3$ $3$ For Question $4$ , we need to apply a line integral to evaluate $3x^2yz^2 - x \sin y + 5xy = 3$ $4$ over the line segment $3x^2yz^2 - x \sin y + 5xy = 3$ $5$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $6$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $7$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $8$ . First, find $3x^2yz^2 - x \sin y + 5xy = 3$ $9$ , which is the differential arc length: $x$ $0$ $x$ $1$ $x$ $2$ Now, set up the line integral: $x$ $3$ Simplify the integrand: $x$ $4$
Homogeneous Differential Equation Solution: For Question $2$ , we need to use implicit differentiation to find $(\frac{\partial z}{\partial x})$ and $(\frac{\partial z}{\partial y})$ from the equation $3x^2yz^2 - x \sin y + 5xy = 3$ . Differentiating with respect to $x$ , we get: $6xyz^2 + 3x^2y(\frac{dz}{dx})z - \sin y + 5y = 0$ Now, solve for $(\frac{dz}{dx})$ . $(\frac{dz}{dx}) = \frac{\sin y - 5y - 6xyz^2}{3x^2yz}$ Differentiating with respect to $y$ , we get: $3x^2z^2 + 3x^2z(\frac{dz}{dy})z - x \cos y + 5x = 0$ Now, solve for $(\frac{dz}{dy})$ . $(\frac{\partial z}{\partial y})$ $0$ For Question $3$ , we need to apply a double integral to evaluate the function $(\frac{\partial z}{\partial y})$ $1$ over the region bounded by $(\frac{\partial z}{\partial y})$ $2$ , $(\frac{\partial z}{\partial y})$ $3$ , $(\frac{\partial z}{\partial y})$ $4$ , and $(\frac{\partial z}{\partial y})$ $5$ . Set up the double integral as: $(\frac{\partial z}{\partial y})$ $6$ First, integrate with respect to $y$ : $(\frac{\partial z}{\partial y})$ $8$ Now, evaluate the inner integral: $(\frac{\partial z}{\partial y})$ $9$ Simplify the expression: $3x^2yz^2 - x \sin y + 5xy = 3$ $0$ Finally, integrate with respect to $x$ : $3x^2yz^2 - x \sin y + 5xy = 3$ $2$ Evaluate the definite integral: $3x^2yz^2 - x \sin y + 5xy = 3$ $3$ For Question $4$ , we need to apply a line integral to evaluate $3x^2yz^2 - x \sin y + 5xy = 3$ $4$ over the line segment $3x^2yz^2 - x \sin y + 5xy = 3$ $5$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $6$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $7$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $8$ . First, find $3x^2yz^2 - x \sin y + 5xy = 3$ $9$ , which is the differential arc length: $x$ $0$ $x$ $1$ $x$ $2$ Now, set up the line integral: $x$ $3$ Simplify the integrand: $x$ $4$ For Question $5$ , we need to solve the homogeneous differential equation $x$ $5$ , with the initial condition $x$ $6$ . Separate variables and integrate: $x$ $7$ $x$ $8$
Homogeneous Differential Equation Solution: For Question $2$ , we need to use implicit differentiation to find $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ from the equation $3x^2yz^2 - x \sin y + 5xy = 3$ . Differentiating with respect to $x$ , we get: $6xyz^2 + 3x^2y\frac{dz}{dx}z - \sin y + 5y = 0$ Now, solve for $\frac{dz}{dx}$ . $\frac{dz}{dx} = \frac{\sin y - 5y - 6xyz^2}{3x^2yz}$ Differentiating with respect to $y$ , we get: $3x^2z^2 + 3x^2z\frac{dz}{dy}z - x \cos y + 5x = 0$ Now, solve for $\frac{dz}{dy}$ . $\frac{\partial z}{\partial y}$ $0$ For Question $3$ , we need to apply a double integral to evaluate the function $\frac{\partial z}{\partial y}$ $1$ over the region bounded by $\frac{\partial z}{\partial y}$ $2$ , $\frac{\partial z}{\partial y}$ $3$ , $\frac{\partial z}{\partial y}$ $4$ , and $\frac{\partial z}{\partial y}$ $5$ . Set up the double integral as: $\frac{\partial z}{\partial y}$ $6$ First, integrate with respect to $y$ : $\frac{\partial z}{\partial y}$ $8$ Now, evaluate the inner integral: $\frac{\partial z}{\partial y}$ $9$ Simplify the expression: $3x^2yz^2 - x \sin y + 5xy = 3$ $0$ Finally, integrate with respect to $x$ : $3x^2yz^2 - x \sin y + 5xy = 3$ $2$ Evaluate the definite integral: $3x^2yz^2 - x \sin y + 5xy = 3$ $3$ For Question $4$ , we need to apply a line integral to evaluate $3x^2yz^2 - x \sin y + 5xy = 3$ $4$ over the line segment $3x^2yz^2 - x \sin y + 5xy = 3$ $5$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $6$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $7$ , $3x^2yz^2 - x \sin y + 5xy = 3$ $8$ . First, find $3x^2yz^2 - x \sin y + 5xy = 3$ $9$ , which is the differential arc length: $x$ $0$ $x$ $1$ $x$ $2$ Now, set up the line integral: $x$ $3$ Simplify the integrand: $x$ $4$ For Question $5$ , we need to solve the homogeneous differential equation $x$ $5$ , with the initial condition $x$ $6$ . Separate variables and integrate: $x$ $7$ $x$ $8$ Apply the initial condition $x$ $6$ to find $6xyz^2 + 3x^2y\frac{dz}{dx}z - \sin y + 5y = 0$ $0$ . This is not possible because $6xyz^2 + 3x^2y\frac{dz}{dx}z - \sin y + 5y = 0$ $1$ is undefined at $6xyz^2 + 3x^2y\frac{dz}{dx}z - \sin y + 5y = 0$ $2$ .