APSC 172 Final Exam Student Workbook Edited By: James Xie and Taylor Sawadsky

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APSC 172 Final Exam Student Workbook Edited by: James Xie and Taylor Sawadsky

Contents Section 1: Vector Forms of Lines and Planes ...... 1 Section 2: Multivariable Differentiation ...... 1 2.1 Level Curves ...... 2 2.2 Gradients ...... 2 2.3 Directional Derivatives ...... 5 Section 3: Application of Multivariable Differentiation...... 7 3.1 Linearization & Tangents to Intersections ...... 7 3.2 Optimization ...... 8 Section 4: Sequences and Series ...... 10 4.1 Properties of Sequences and Series ...... 10 4.2 Geometric Series: ...... 11 4.3 Alternating Series ...... 12 4.4 Power Series...... 13 Section 5: Taylor Series ...... 14 Section 6: Multiple Integrals ...... 17 6.1 Double Integrals ...... 17 6.2 Polar Coordinates ...... 20 6.3 Centre of Mass ...... 21 6.4 Triple Integrals ...... 22

Section 1: Vector Forms of Lines and Planes

Lines may be fully described by a starting point (풗ퟎ) and a direction (풖):

풗 = 풗ퟎ + 푡풖

A plane may be using the vector dot product – a multiplication operation performed on a pair of vectors (of the same dimensionality) which always produces a scalar.

[푎, 푏, 푐] ∙ [푥, 푦, 푧] = 푎푥 + 푏푦 + 푐푧

Alternatively, it can be expressed in scalar form as:

풖 ∙ 풗 = ‖풖‖‖풗‖ cos 휃 ⇒ If 풖 ⊥ 풗, 풖 ∙ 풗 = 0

Where ||v|| represents the magnitude of the vector, given by:

2 2 2 ||푣1, 푣2, 푣3|| = √푣1 + 푣2 + 푣3

A plane can therefore be described as the set of all points (x, y, z) from a reference (x0, y0, z0) such that the vector between them is perpendicular to another vector to (a, b, c):

[(푥 − 푥0), (푦 − 푦0), (푧 − 푧0)] ∙ 풏 = 0 Section 2: Multivariable Differentiation A partial derivative (휕) is the derivative of a function with respect to a variable, while holding all others constant. The rules for implicit differentiation are the same, with the modified notation.

Given a function z = f(x, y), a partial derivative of z may be done with respect to x or y:

With respect to With respect to y, 휕푧 x, 휕푧 x is treated as a = 푓 (푥, 푦) = 푓푦(푥, 푦) 휕푥 푥 y is treated as a 휕푦 constant constant Geometrically, a partial derivative is the slope of the graph in a direction.

Second derivatives are similarly performed:

휕2푧 휕2푧 휕2푧 휕2푧 = 푓 (푥, 푦) = 푓 (푥, 푦) = 푓 (푥, 푦) = 푓 (푥, 푦) 휕푥2 푥푥 휕푦2 푦푦 휕푥휕푦 푥푦 휕푦휕푥 푦푥 Differentiation on smooth curves is commutative – it does not matter the order of differentiation:

휕 휕푧 휕 휕푧 휕2푧 ( ) = ( ) = ⟺ 푓 (푥, 푦) = 푓 (푥, 푦) 휕푥 휕푦 휕푦 휕푥 휕푦휕푥 푥푦 푦푥

The chain rule is also extended to describe the total derivative of z with respect to an external parameter t, and implies that x and y are purely functions of t.

푑푧 휕푧 푑푥 휕푧 푑푦 = + , 푧 = 푓(푥(푡), 푦(푡)) 푑푡 휕푥 푑푡 휕푦 푑푡

2.1 Level Curves A level curve is a family of points which produce the same value in the function. The level curves on a surface form contour lines – following a level curve results in no change in “height”.

A function z = f(x, y) treats z as dependent on (x, y), but can be generalized into a function F:

퐹 = 푓(푥, 푦) − 푧 where the original function is a level curve at F = 0.

2.2 Gradients The gradient of a function (훁푓) is a vector containing its partial derivatives, which points in the direction of largest slope – i.e. perpendicularly to the level curves of the surface. The magnitude of the gradient is the size of the slope. 휕푓 휕푓 휕푓(풙⃑⃑ ) 훁푓 = ( , ) = (푓 , 푓 ) = 휕푥 휕푦 푥 푦 휕풙⃑⃑ Memorize This! ||훁푓|| = slope

Note in this example, f(x,y) produces a gradient vector which is on the xy plane.

Example 2.2.1 Let 퐹(푥, 푦, 푧) = (푦 + 푧)푒푧−푥, suppose the equation 퐹(푥, 푦, 푧) = 3 defines 푧 as a function of 푥 and 푦 implicity in the neighbourhood of the point (1,2,1). Find the equation of the tangent plane to the level surface of F at the point (1,2,1) using two different methods: a) The gradient of F b) The gradients of (푦 + 푧) 푒푧−푥 = 3

2.3 Directional Derivatives

The directional derivative of a function (퐷푢푓) at a point (x0, y0, z0) is its slope in a specified direction (u=(ux,uy,uz)) and is the amount of the gradient which is parallel to the direction: 훁푓(푥 , 푦 , 푧 ) ∙ 풖 퐷 푓(푥 , 푦 , 푧 ) = 0 0 0 푢 0 0 0 ||풖|| Memorize This! 푓 (푥 , 푦 , 푧 )푢 + 푓 (푥 ,푦 , 푧 )푢 + 푓 (푥 ,푦 , 푧 )푢 = 푥 0 0 0 푥 푦 0 0 0 푦 푧 0 0 0 푧 2 2 2 √푢푥 + 푢푦 + 푢푧

From the dot product, if 풖 ⊥ 훁푓, then 퐷푢푓 = 0, and it is maximized (but not equal to) when they are in the same direction. This equation can also be adapted to work in two dimensions.

The slope (rate of increase) is equivalent to the directional derivative travelling at unit speed:

vertical velocity 푧̇ 푑푥 푚 = = , 푥̇ = horizontal speed ||푥̇ + 푦̇|| 푑푡

Example 2.3.1

Let the temperature at a point (x, y, z) in space be defined by 푇(푥, 푦, 푧) = 푥푦푧.

a) In what direction is T increasing most rapidly at the point (2, 1, 1) and what is this maximum rate of increase?

b) Calculate the directional derivative of T(2, 1, 1) in the direction of the vector [3, 0, 4]

c) A curve C is formed at the intersection of the surface 푥2 + 4푦2 − 2푧5 = 6 with the plane 4푥 + 3푦 − 5푧 = 6. A bug at the point (2, 1, 1) is moving along C at speed 12. What rate is it experiencing the temperature change (using 푇 = 푥푦푧 as above)?

Section 3: Application of Multivariable Differentiation 3.1 Linearization & Tangents to Intersections The linear approximation of a function at a point in ℝ3 is a tangent plane to a surface. For a point

(x0, y0, z0) on a surface z = f(x, y), the linearization will have the same gradient as the surface at the point:

(푧 − 푧0) = ∇푓(푥0, 푦0) ∙ [(푥 − 푥0), (푦 − 푦0)]

휕푧 휕푧 = | (푥 − 푥0) + | (푦 − 푦0) 휕푥 ( ) 휕푦 푥0,푦0 (푥0,푦0)

Alternatively, a normal vector to the plane may be used which corresponds to the gradient at the Gpoint on the level surface F(x, y, z):

[(푥 − 푥0), (푦 − 푦0), (푧 − 푧0)] ∙ ∇퐹(푥0, 푦0, 푧0) = 0

휕푧 휕푧 | (푥 − 푥 ) + | (푦 − 푦 ) − (푧 − 푧 ) = 0 휕푥 0 휕푦 0 0 (푥0,푦0) (푥0,푦0)

The error of the linear approximation is given by: 휕푧 휕푧 ∆푧 = | ∆푥 + | ∆푦 휕푥 ( ) 휕푦 푥0,푦0 (푥0,푦0)

In the intersection of two surfaces f and g, the tangent line to the intersection at a point (x0, y0, z0) is perpendicular to both gradients of the level curves F and G at that point:

[푥, 푦, 푧] = [푥0, 푦0, 푧0] + 푡풅, 풅 ∙ 훁퐹 = 풅 ∙ 훁퐺 = 0 Example 3.1.1

Given 퐴(푥, 푦, 푧) = 푥2 − 3푦2 + 4푧 and 퐵(푥, 푦, 푧) = −푥2 + 푦 − 2푧2 + 4, write the equation of the line tangent to the curve of intersection of A and B at point (3, 3, 2).

3.2 Optimization In multivariable functions, local maxima/minima occur if all 1st partial derivatives are 0 at the point and if the function is decreasing/increasing in all directions. If only the first condition is met, it is a saddle point.

휕푓 휕푓 | = | = 0 휕푥 (푎,푏) 휕푦 (푎,푏)

The second derivative test evaluates the determinant of the hessian matrix – this is only for functions of two variables. 퐷 > 0 and 푓 , 푓 > 0 ⟹ (푎, 푏) is a minima 푥푥 푦푦

퐷 > 0 and 푓 , 푓 < 0 ⟹ (푎, 푏) is a maxima 2 푥푥 푦푦 퐷(푎,푏) = 푓푥푥(푎, 푏)푓푦푦(푎, 푏) − [푓푥푦(푎, 푏)] , 퐷 < 0 ⟹ (a, b) is a saddle point

{ 퐷 = 0 ⟹ the test fails

In the cases where D > 0, neither fxx or fyy can be equal to 0.

Example 3.2.1

Find the critical points of the function 푓(푥, 푦) = (푥2 + 푦2)푒−푥 and decide in each case whether it is a maximum, minimum, or saddle point.

Point 푓푥푥 푓푦푦 푓푥푦 퐷 Result

Section 4: Sequences and Series 4.1 Properties of Sequences and Series

A sequence is any list of numbers {푎푛}, while a series (S) is the sum of a sequence: ∞

푆 = ∑ 푎푛 = 푎1 + 푎2 + 푎3 + ⋯ 푛=1

The nth partial sum (푠푛) of a series is the sum of its first n terms: 푛

푠푛 = ∑ 푎푛 = 푎1 + 푎2 + 푎3 + ⋯ + 푎푛 푛=1

If the sequence of partial sums {푠푛} converges to a finite limit 푆 as 푛 approaches infinity, the series is convergent and 푆 is the sum of the series: ∞

푆 = ∑ 푎푛 = lim 푠푛 푛→∞ 푛=1

This requires the sequence {푎푛} to converge to 0. If a series does not converge, it is divergent. Example 4.1.1 By any method find an expression for the sum of the infinite series and simplify your expression as much as possible. ∞ 푥푛+1 푥2 푥3 푥4 푥5 푓(푥) = ∑ = + + + + ⋯ 푛(푛 + 1) 1 ∙ 2 2 ∙ 3 3 ∙ 4 4 ∙ 5 푛=1

Hint: ∫ ln(푥) 푑푥 = 푥 ln(푥) − 푥 + 푐

4.2 Geometric Series: Each term in a geometric series is solved from the previous term by multiplying by a constant 푟. ∞ ∑ 푎푟푛−1 = 푎 + 푎푟 + 푎푟2 + 푎푟3 + ⋯ 푛=1 푎 = ⟺ |푟| < 1 1 − 푟

푎(1 − 푟푛) 푠 = 푛 1 − 푟

These series converge if and only if |푟| < 1 If r is a negative number, the series becomes alternating: ∞ ∑(−푟)푛−1 = 1 − 푟 + 푟2 − 푟3 + ⋯ 푛=1 4.3 Alternating Series Each term in an alternating series follows some sequence rule, but also changes signs each time.

∞ 푛+1 ∑(−1) ∙ 푏푛 = 푏1 − 푏2 + 푏3 − 푏4 + ⋯ , 푏푛 > 0 푛=1

Alongside the general rule for convergence, the alternating series test states that if 푏푛+1 < 푏푛 and lim 푏푛 = 0, then the series will also converge. 푛→∞ The nth partial sum of the series may be used as an approximation for the series. The error of the th n partial sum is less than bn+1:

Error = |푆 − 푠푛| ≤ 푏푛+1

Example 4.3.1 Find the sum of the series correct to 3 decimal places. (ie. with error < 0.001) (−1)푛−1 푆 = ∑ 푛!

4.4 Power Series From the generic geometric series, if we substitute r with x:

∞ 푎 푆 = ∑ 푎푟푛 = 1 − 푟 푛=0 푎 Thus, the function now has a series representation (for a certain domain of r). Other functions 1−푟 can be represented with different r(x) as well as manipulating the series by addition/subtraction/scaling and integration/differentiation.

The interval of convergence describes the possible values of x where the series representation is valid. It follows from the initial interval of |r| < 1.

Example 4.4.1: Express 4 as the sum of a power series and find the interval of convergence. 1+푥2

Section 5: Taylor Series A function can be approximated nearby a point using its derivatives, called its Taylor Series Expansion. This is an extension of linearization to include higher-order derivatives. The nth partial sum of a Taylor series is a Taylor polynomial. If the point is at the origin, the approximation is a MacLauren Series.

∞ 1 1 1 푓(푥) ≅ ∑ 푓(푛)(푎)(푥 − 푎)푛 = 푓(푎) + 푓′(푎)(푥 − 푎) + 푓(2)(a)(x-a)2+ f (3)(푎)(푥 − 푎)3 + ⋯ 푛! 2! 3! 푛=0 Taylor series do not necessarily converge to the function as n approaches infinity.

푛 1 푇 (푥)| = ∑ 푓(푛)(푎)(푥 − 푎)푛 푛 푎 푛! 푛=0

Example 5.1 Write the Taylor series expansion of 푓(푥) = sin 푥 at x = 0 to the first 5 non-zero terms and use it to approximate x = 2 and x = 3. Use the Taylor Error Estimate to find the error in both cases. 푛 푓(푛)(푥) 푓(푛)(푎) 1/푛! 0 1 2 3 4 5 6 7 8

9 10

For x in an interval [a – d, a + d] on an nth degree Taylor polynomial centered at x = a, the error is no greater than:

‖푓(푛+1)(푥)‖ 퐸 ≤ 푑푛+1 푛 (푛 + 1)!

Where ‖푓(푛+1)(푥)‖ is the maximum value the function (푓(푛+1)(푥)) attains on the given interval.

Example 5.2 Use Taylor’s inequality to decide how many terms of the Taylor series are needed to approximate ex centered at x = 0 to within 10-6 between the interval [-1, 1]:

Some important MacLauren Series:

∞ 1 Memorize = ∑ 푥푛 = 1 + 푥 + 푥2 + 푥3 + ⋯ 1 − 푥 This! 푛=0

∞ 푥푛 푥 푥2 푥3 푥4 푒푥 = ∑ = 1 + + + + + ⋯ 푛! 1! 2! 3! 4! 푛=0

푘푥2 푘푥3 (1 + 푥)푘 = 1 + 푘푥 + (푘 − 1) + (푘 − 1)(푘 − 2) + ⋯ 2! 3!

푥3 푥5 푥7 sin 푥 = 푥 − + − + ⋯ 3! 5! 7!

푥2 푥4 푥6 cos 푥 = 1 − + − + ⋯ 2! 4! 6! Section 6: Multiple Integrals 6.1 Double Integrals Multiple integrals can be broken into successive single integrals evaluated from inside outwards.

When integrating along one direction, treat all other variables as constants and evaluate the bounds after each integral. Once a variable has been integrated over and evaluated, it should not appear in any further integrals.

푦2 푥2 ∬ 푓(푥, 푦) 푑퐴 = ∫ ∫ 푓(푥, 푦) 푑푥 푑푦

푅 푦1 푥1 The above may be read as the integral of f(x,y) over the region R, which can represent a volume.

The volume of a solid between a surface z1 = f(x, y) and z2 = g(x, y) inside a region R is:

푉 = ∬ 푓(푥, 푦) − 푔(푥, 푦) ∙ 푑퐴 푅 Multiple integrals can be broken into slices perpendicular to an axis of your choice, transforming the double integral into iterated single integrals:

푏 ∬ 푓(푥, 푦) 푑퐴 = ∫ 퐴(푥) 푑푥 푅 푎 푏 퐴(푥) = ∫ (푔(푦) − ℎ(푦)) 푑푦 푎 Example 6.1.1

Find the volume of the solid that lies under the plane 푧푇 = 푥 + 2푦 and above the region 퐷 in the x-y plane bounded by the line 푦 = 2푥 and the parabola 푦 = 푥2.

Example 6.1.2 Find the volume of the tetrahedron bounded by the planes: 푧 = 2 − 푥 − 2푦, 푥 = 2푦, 푥 = 0, 푧 = 0

Example 6.1.3 Evaluate the following iterated integral by interchanging the order of integration. 1 1 ∫ ∫ 푦2 푑푦 푑푥 0 푥

6.2 Polar Coordinates Cartesian coordinates describe points linearly (x, y) and define areas as rectangular regions 퐴 = ∆푥 ∙ ∆푦.

Each point can also be described as a distance from the origin (modulus) and angle from the horizontal (argument), (푟, 휃).

푥 = 푟 cos 휃 푦 = 푟 sin 휃 푥2 + 푦2 = 푟2 Areas are now defined as 퐴 = 푟∆휃 ∙ ∆푟 since the arc length changes as radius increases:

Example 6.2.1:

Use polar coordinates to evaluate the integral

∬(3푥 + 4푦2) ∙ 푑퐴 푅 where 푅 is the region in the upper half plane bounded by 푥2 + 푦2 = 1 and 푥2 + 푦2 = 4. 1 (Tech help: ∫ sin2 휃푑휃 = (푥 − sin 휃 cos 휃)) 2

6.3 Centre of Mass Given an extended object V with a density p, its center of mass (푥̅) is the weighted average of each individual differential volume dV. For linear objects:

1 푏 푥 = ∫ 푥 휌(푥) ∙ 푑푥 푀 푎 If the mass is distributed over a planar region 푅, the the centroid will have 2 components (푥, 푦) 1 1 푥 = ∬ 푥 휌(푥, 푦) ∙ 푑퐴, 푦 = ∬ 푦 휌(푥, 푦) ∙ 푑퐴 푀 푀 푅 푅

Example 6.3.1 Find the mass of a triangle with vertices (0,0), (1,0), and (0,2) and a density function 휌(푥, 푦) = 푥, and calculate the location of its center of mass.

6.4 Triple Integrals Extending to functions of 3 variables, the integration now occurs over solid regions in ℝ3. The same principles apply.

푧2 푦2 푥2 ∭ 푓(푥, 푦, 푧) 푑푉 = ∫ ∫ ∫ 푓(푥, 푦, 푧) 푑푥 푑푦 푑푧

퐸 푧1 푦1 푥1 If the function f(x,y,z) is 1, the triple integral simply calculates the volume.

Example 6.4.1 퐸 is the tetrahedron bounded by 4 planes, 푥 = 0, 푦 = 0, 푧 = 0, 푥 + 푦 + 푧 = 1. Find its mass and find 푥̅ and 푦̅ if it’s density is 휌(푧) = 푧.

Example 6.4.2 Change the order of integration to 푑푥 푑푦 푑푧 for the integral and evaluate. 1 1 1−푦 ∫ ∫ ∫ 푑푧 푑푦 푑푥

0 √푥 0