Math 251 Exam 2 Review PDF

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Rosanna Pearlstein Exam 2 Review...

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Math 251 Exam 2 Review

Format: The exam will have 2 Free Response Questions and 8 Multiple Choice Questions, and the rest will be multiple choice on an 882-E scantron. The Exam will cover sections 14.1 to 15.3 except 14.2 and 14.8. Some sketching is required in 2-D.

Section 14.1 – Functions of several Variables: This section is pretty straight forward. It’s all about sketching the domain as well as the level curves of a function of multiple variables. To sketch the domain, all you need to do is figure out the continuity of the functions and by using the rules of sketching inequalities figure out what is included and what isn’t included. Definition of Graphs: If f is a function of two variables with domain D, then the graph of f is the set of all points (x,y,z) in R3 such that z = f(x,y) and (x,y) is in D. Definition of Level Curves: The Level Curves of a function f of two variables are the curves with equations f(x,y) = k, where k is a constant (in the range of f). To sketch a level curve, you pick the values of k and plug into the equation and sketching those equations. When dealing with functions of three or more variables, the sketching of the domain and level surfaces will be in 3D and will be most likely a plane.

Section 14.3 – Partial Derivatives: This section deals with partial derivatives which is taking the derivative of multivariable function with respect to one variable while making the other variable a constant. Higher Derivatives: You can take the derivative again (just as Cal 1) but with respect to other variables. For example you can take the partial derivative of a function with respect to x and then take the partial

derivative of that partial derivative with respect to y. Clairaut’s Theorem: Suppose f is defined on a disk D that contains the point (a,b). If the functions fxy and fyx are both continuous on d, then 𝑓𝑥𝑦 (𝑎, 𝑏) = 𝑓𝑦𝑥 (𝑎, 𝑏) 𝜕2𝑢

Laplace’s Equation: 𝜕𝑥2 + 𝜕2𝑢

Wave Equation: 𝜕𝑡 2 = 𝑎 2

𝜕2𝑢

𝜕𝑦 2

=0

𝜕2𝑢 𝜕𝑥 2

The trickiest part of this section is going to be actually differentiating the equations, meaning getting the derivatives right for certain functions. There will be a list of common derivatives below.

Section 14.4 – Tangent Planes and Linear Approximations: The section piggybacks off of some of the sections from the first exam. This section involves the tangent plane as well as the equations for the tangent planes. That equation is as follows: 𝑧 − 𝑧𝑜 = 𝑓𝑥 (𝑥𝑜 , 𝑦𝑜 )(𝑥 − 𝑥 𝑜 ) + 𝑓𝑦 (𝑥𝑜 , 𝑦𝑜 )(𝑦 − 𝑦𝑜 ), where fx and fy are the partial derivatives of the equation of the surface given, and (xo, yo, zo) is a point given to you.

Linear Approximations: 𝐿(𝑥, 𝑦) = 𝑓(𝑎, 𝑏) + 𝑓𝑥 (𝑎, 𝑏)(𝑥 − 𝑎) + 𝑓𝑦 (𝑎, 𝑏)(𝑦 − 𝑏), where (x,y) is the point at which you are approximating, and (a,b) is the point you are given in the very beginning.

Theorem: If the partial derivatives fx and fy exist near (a,b) and are continuous at (a,b), then f is differentiable at (a,b). L is equal to the linearization of the function. Differential: 𝑑𝑧 = 𝑓𝑥 (𝑎, 𝑏 )(𝑥 − 𝑎 ) + 𝑓𝑦 (𝑎, 𝑏 )(𝑦 − 𝑏 ) 𝑜𝑟 𝑑𝑧 = 𝑓𝑥 𝑑𝑥 + 𝑓𝑦 𝑑𝑦 This section can be a little tricky but we will use it to estimate a value

Section 14.5 – The Chain Rule: This section deals with the chain rule of a function when there are multiple variables. You don’t really need to memorize anything from this section except for this equation 𝑑𝑧 𝜕𝑓 𝑑𝑥 𝜕𝑓 𝑑𝑦 Case 1: = + 𝑑𝑡 . Case 2 is derived from the Case 1 equation except it involves more 𝑑𝑡

𝜕𝑥 𝑑𝑡

𝜕𝑦

than one additional variable (Ex: (s,t)).

Implicit Differentiation: 𝜕𝐹 𝑑𝑥

𝜕𝐹 𝑑𝑦 =0 𝜕𝑥 𝑑𝑥 𝜕𝑦 𝑑𝑥 𝜕𝐹 𝜕𝑧 𝜕𝑥 =− 𝜕𝐹 𝜕𝑥 𝜕𝑧 𝜕𝐹 𝜕𝑧 𝜕𝑦 =− 𝜕𝐹 𝜕𝑦 𝜕𝑧 +

This section can be extremely tricky due to the format of the Chain Rule equation. To help relieve this difficulty, the best way to do it is by making a chart.

Section 14.6 – Directional Derivatives and the Gradient Vector: The directional derivative of f at (xo, yo) in the direction of a unit vector u = is equal to the slope. 𝐷𝑢 𝑓(𝑥, 𝑦) = 𝑓𝑥 (𝑥, 𝑦 )𝑎 + 𝑓𝑦 (𝑥, 𝑦 )𝑏 = 𝑓𝑥 (𝑥, 𝑦 )𝑐𝑜𝑠∅ + 𝑓𝑦 (𝑥, 𝑦)𝑠𝑖𝑛∅ The tricky part of this is when you are given a vector that isn’t a unit vector. In this case, you will have to transform the vector into a unit vector. The gradient vector is the vector of the partial derivatives of a function f. It is given by ∇𝑓(𝑥, 𝑦 ) = 〈𝑓𝑥 (𝑥, 𝑦 ), 𝑓𝑦 (𝑥, 𝑦)〉. The gradient vector and directional derivative also applies for 3 variable equations. Theorem: Suppose f is differentiable of two or three variables. The maximum value of the directional derivative 𝐷𝑢 𝑓(𝑥) is |∇𝑓(𝑥)|, and it occurs when u has the same direction as the gradient vector ∇𝑓(𝑥). Tangent Planes to Level Surfaces: 𝐹𝑥 (𝑥𝑜 , 𝑦𝑜 , 𝑧𝑜 )(𝑥 − 𝑥𝑜 ) + 𝐹𝑦 (𝑥𝑜 , 𝑦𝑜 , 𝑧𝑜 )(𝑦 − 𝑦𝑜 ) + 𝐹𝑧 (𝑥𝑜 , 𝑦𝑜 , 𝑧𝑜 )(𝑧 − 𝑧𝑜 ) = 0 ∇𝐹(𝑝𝑜 ) ∙ 𝑝𝑜 𝑝 = 0 This section is pretty easy just a lot of formulas to memorize.

Section 14.7 – Maximum and Minimum Values: This section is based off of Calculus 1 but with multiple variables. Definition: A function of two variables has a local maximum at (a,b) if f(x,y) ≤ f(a,b) when (x,y) is near (a,b). [This means that f(x,y) ≤ f(a,b) for all points (x,y) in some disk with center (a,b).] The number f(a,b) is called a local maximum value. If f(x,y) ≥ f(a,b) when (x,y) is near (a,b), then f has a local minimum at (a,b) and f(a,b) is a local minimum value. Theorem: If f has a local maximum or minimum at (a,b) and the first-order partial derivatives of f exist there, then fx(a,b) = 0 and fy(a,b) = 0. (Critical Points). Second Derivative Test: Suppose the second partial derivatives of f are continuous on a disk with center (a,b), and suppose that fx(a,b) = 0 and fy(a,b) = 0 [that is, (a,b) is a critical point f]. Let 𝐷 = 𝐷(𝑎, 𝑏) = 𝑓𝑥𝑥 (𝑎, 𝑏 )𝑓𝑦𝑦 (𝑎, 𝑏) − [𝑓𝑥𝑦 (𝑎, 𝑏)]2 If D > 0 and fxx(a,b) > 0, then f(a,b) is a local minimum. If D > 0 and fxx(a,b) < 0, then f(a,b) is a local maximum. If D < 0, then f(a,b) is not a local maximum or minimum, and f(a,b) is a saddle point. Extreme Value Theorem for Functions of Two Variables: If f is continuous on a closed, bounded set d in R2, then f attains an absolute maximum value f(x1, y1) and an absolute minimum value f(x2, y2) at some points (x1, y1) and (x2,y2) in D. Steps to find the absolute Maximum and Minimum Values: 1.) Find the values of f at the critical points of f in D. 2.) Find the extreme values of f on the boundary of D. 3.) The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value. This section will be tricky and requires a lot of practice to understand it. It is very similar to that of Calculus 1 but just in 3D.

Section 15.1 – Double Integrals over Rectangles: This section is pretty straight forward and deals with taking a double integral over a function of two variables with respect to x and y. 𝑉 = ∬ 𝑓(𝑥, 𝑦)𝑑𝐴 When f factors as a product of two functions g and h, the integral splits as a product. 𝑏

𝑑

∫ 𝑔(𝑥)𝑑𝑥 ∫ ℎ(𝑦)𝑑𝑦 𝑐

𝑎

Average Value Formula: 1 ∬ 𝑓(𝑥, 𝑦)𝑑𝐴 𝐴(𝑅) This section is pretty easy the only confusing part will be messing up the integrals. This section is the building block for 15.2 and 15.3.

Section 15.2 – Double Integrals over General Regions: This section is a little trickier because there are multiple cases and you have to figure out which case applies. Type 1: 𝐷 = {(𝑥, 𝑦)|𝑎 ≤ 𝑥 ≤ 𝑏, 𝑔1 (𝑥) ≤ 𝑦 ≤ 𝑔2 (𝑥)} Basically, the range (y values) are given by two equations with one on top of each other. The formula for this is: 𝑏

∫ ∫ 𝑎

𝑔2 (𝑥)

𝑔1 (𝑥)

𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑥

Type 2: 𝐷 = {(𝑥, 𝑦)|𝑐 ≤ 𝑦 ≤ 𝑑, ℎ1 (𝑦) ≤ 𝑥 ≤ ℎ 2 (𝑦) }

Basically, the domain (x values) are between two functions with respect to y. 𝑑

∫ ∫ 𝑐

ℎ2 (𝑦)

ℎ1 (𝑦)

𝑓(𝑥, 𝑦 )𝑑𝑥𝑑𝑦

Properties of Double Integrals: ∬[𝑓(𝑥, 𝑦) + 𝑔 (𝑥, 𝑦 )]𝑑𝐴 = ∬ 𝑓(𝑥, 𝑦 )𝑑𝐴 + ∬ 𝑔(𝑥, 𝑦 )𝑑𝐴 𝐷

𝐷

∬ 𝑐𝑓(𝑥, 𝑦)𝑑𝐴 = 𝑐 ∬ 𝑓(𝑥, 𝑦)𝑑𝐴 𝐷

𝐷

𝐷

∬ 𝑓(𝑥, 𝑦) ≥ ∬ 𝑔(𝑥, 𝑦) 𝐷

𝐷

∬ 1𝑑𝐴 = 𝐴(𝐷) 𝐷

This section can get difficult when you need to switch the bounds to solve an impossible integral. But will become easier with practice.

Section 15.3 – Double Integrals in Polar Coordinates: 𝑟2 = 𝑥2 + 𝑦 2 𝑥 = 𝑟𝑐𝑜𝑠∅ 𝑦 = 𝑟𝑠𝑖𝑛∅ This Section involves turning equations into polar coordinates as well as the bounds. 𝛽

𝑏

∬ 𝑓(𝑥, 𝑦 )𝑑𝐴 = ∫ ∫ 𝑓(𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃)𝑟𝑑𝑟𝑑𝜃 𝑅

𝛼

𝑎

This section is pretty straight forward, but the hardest part is going to be figuring out what theta is and how to find theta....