Calc 3 Practice Problems

Multivariable Calculus Practice Problems: A Comprehensive Guide

Multivariable calculus extends the concepts of single-variable calculus to functions of multiple variables, introducing new tools and techniques to analyze and optimize these functions. This guide provides a diverse set of practice problems, covering key topics such as partial derivatives, multiple integrals, vector calculus, and applications. Each problem is designed to reinforce understanding and build problem-solving skills.

Section 1: Partial Derivatives

Partial derivatives allow us to measure how a function changes with respect to one variable while holding others constant.

Problem 1:
Given the function ( f(x, y) = x^3y^2 + 2xy^3 ), find ( f_x(x, y) ) and ( f_y(x, y) ).

Solution:
[ f_x(x, y) = \frac{\partial}{\partial x}(x^3y^2 + 2xy^3) = 3x^2y^2 + 2y^3 ] [ f_y(x, y) = \frac{\partial}{\partial y}(x^3y^2 + 2xy^3) = 2x^3y + 6xy^2 ]

Problem 2:
For ( g(x, y, z) = e^{x^2y} \sin(z) ), find ( g_{xy}(x, y, z) ).

Solution:
First, find ( g_x ):
[ gx = \frac{\partial}{\partial x}(e^{x^2y} \sin(z)) = 2xy e^{x^2y} \sin(z) ]
Then, find ( g{xy} ):
[ g_{xy} = \frac{\partial}{\partial y}(2xy e^{x^2y} \sin(z)) = 2x e^{x^2y} \sin(z) + 2x(2xy^2 e^{x^2y} \sin(z)) = 2x e^{x^2y} \sin(z)(1 + 2xy^2) ]

Section 2: Multiple Integrals

Multiple integrals generalize the definite integral to functions of more than one variable.

Problem 3:
Evaluate ( \iint_R (x^2 + y^2) \, dA ), where ( R ) is the region bounded by the circle ( x^2 + y^2 = 4 ).

Solution:
Convert to polar coordinates: ( x = r \cos \theta ), ( y = r \sin \theta ), ( dA = r \, dr \, d\theta ).
The bounds are ( 0 \leq r \leq 2 ) and ( 0 \leq \theta \leq 2\pi ).
[ \iint_R (x^2 + y^2) \, dA = \int_0^{2\pi} \int_0^2 r^2 \cdot r \, dr \, d\theta = \int_0^{2\pi} \int_0^2 r^3 \, dr \, d\theta ] [ = \int_0^{2\pi} \left[ \frac{r^4}{4} \right]_0^2 \, d\theta = \int_0^{2\pi} 4 \, d\theta = 8\pi ]

Problem 4:
Set up the triple integral to find the volume of the solid bounded by ( z = x^2 + y^2 ) and ( z = 4 ).

Solution:
The projection of the solid onto the ( xy )-plane is the circle ( x^2 + y^2 = 4 ).
[ V = \iiint_E dV = \int_0^{2\pi} \int0^2 \int{r^2}^4 r \, dz \, dr \, d\theta ]

Section 3: Vector Calculus

Vector calculus deals with vector fields and operations like gradient, divergence, and curl.

Problem 5:
Given ( \mathbf{F}(x, y, z) = (x^2y, yz, xz) ), compute ( \text{div}(\mathbf{F}) ) and ( \text{curl}(\mathbf{F}) ).

Solution:
[ \text{div}(\mathbf{F}) = \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x^2y) + \frac{\partial}{\partial y}(yz) + \frac{\partial}{\partial z}(xz) = 2xy + z + x ] [ \text{curl}(\mathbf{F}) = \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \ x^2y & yz & xz \end{vmatrix} = (z - z, x - x, 0) = (0, 0, 0) ]

Problem 6:
Verify Green’s Theorem for ( \mathbf{F}(x, y) = (y^2, x^2) ) over the rectangle ( [0, 1] \times [0, 1] ).

Solution:
[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_R \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA ] [ \frac{\partial Q}{\partial x} = 2x, \quad \frac{\partial P}{\partial y} = 2y \quad \Rightarrow \quad \iint_R (2x - 2y) \, dA = 0 ] Line integral calculation confirms the result.

Section 4: Applications

Multivariable calculus has applications in optimization, physics, and engineering.

Problem 7:
Find the maximum volume of a rectangular box with a fixed surface area of 36 square units.

Solution:
Let ( V = xyz ) and ( S = 2(xy + yz + xz) = 36 ).
Use Lagrange multipliers or substitution to maximize ( V ).
The maximum volume occurs when ( x = y = z = 3 ), giving ( V = 27 ).

Problem 8:
Find the directional derivative of ( f(x, y) = x^2y ) at ( (2, 3) ) in the direction of ( \mathbf{v} = \langle 1, 2 \rangle ).

Solution:
[ \nabla f(x, y) = \langle 2xy, x^2 \rangle \quad \Rightarrow \quad \nabla f(2, 3) = \langle 12, 4 \rangle ] [ D_{\mathbf{v}}f = \nabla f \cdot \frac{\mathbf{v}}{|\mathbf{v}|} = \langle 12, 4 \rangle \cdot \left\langle \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right\rangle = \frac{12 + 8}{\sqrt{5}} = \frac{20}{\sqrt{5}} = 4\sqrt{5} ]

Conclusion:
These practice problems cover the core concepts of multivariable calculus, from partial derivatives to vector calculus and applications. Regular practice and conceptual understanding are key to mastering this subject. Use these problems to test your skills and deepen your knowledge!