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RPI - MATH 2010 - Class Notes - Week 2

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MATH 2010 – Multivariable Calculus & Matrix Algebra

Professor Herron – Rensselaer Polytechnic Institute

Week 2 (2/1/16 – 2/5/16)

Important: These notes are in no way intended to replace attendance in lecture. For best results in this course, it is imperative that you attend lecture and take your own detailed notes. Please keep in mind that these notes are written specifically with Professor Herron’s sections in mind, and no one else’s. There were no important pictures or diagrams in lecture this week, so I will not be placing images in this set of notes.

Gradients

- Two kinds of functions

▪ Real value functions

o �� = ��(��, ��) = ��(��⃑)

▪ Vector value functions

o ��⃑ = ��(��); ��⃑ = 〈��1, ��2〉 and ��⃑ = 〈��1, ��2〉

- A gradient is a special type of vector value function, defined by partial derivatives ▪ If �� is a function of two variables, �� and ��, then the gradient of �� is the vector function ∇�� defined by: ∇��(��, ��) = 〈����(��, ��), ����(��, ��)〉 =���� Don't forget about the age old question of What are the hybridization experiments?

���� ��̂+����

���� ��̂

▪ In three dimensions: ∇��(��, ��, ��) = 〈����, ����, ����〉 =����

���� ��̂+����

���� ��̂+����

���� ��̂

Example 1 We also discuss several other topics like Who built the zagwe dynasty?

- �� = ��2 + ��2; find the gradient and the value of ��⃑.

▪ ∇��(��, ��) = 2����̂+ 2����̂= 2��⃑

▪ ��⃑ = ����̂+ ����̂

▪ ‖��⃑⃑‖ = √���� + ���� (the level curves are circles)

iClicker

- A unit vector in the direction of vector ��̂= 〈��, ��, ��〉 has formulae:

▪1‖��⃑‖��⃑

▪1

√��2+��2+��2��⃑

▪1‖��⃑‖〈��, ��, ��〉

Applications of Gradients: 1

- Chain rule for paths If you want to learn more check out What is the meaning of parasitology?

▪ Suppose a particle is moving on a surface �� = ��(��, ��). Its position is (��(��), ��(��), ��(��)). Find its velocity vector. Don't forget about the age old question of What genetic disease causes brittle bones and problems in the liver, spleen, and bone marrow?

▪ ��′(��) =��������(��(��), ��(��)) =���� ����

����

����+���� ����

���� ����

▪ Suppose that �� = ��(��, ��) is a differentiable function of �� and ��, where �� = ��(��) and �� = ℎ(��) are both differentiable forms of ��. Then �� is a differentiable function of �� and ���� Don't forget about the age old question of What are the economic developments after world war 1?

If you want to learn more check out Matter is comprised of what?

����=���� ����

����

����+���� ����

���� ����.

Example 2

- �� = ��((��������(��))2, (��������(��))2). Find ����

▪����

���� = 2��;����

���� = 2��;����

����.

����= cos(��) − ��������(��);����

����= sin(��) + ��������(��)

▪����

����

����

����+���� ����

����

����= 2(��������(��))(cos(��) − ��������(��)) + 2(��������(��))(sin(��) + ��������(��))

▪ = 2��������2(��) − 2��2cos(��) sin(��) + 2��������2(��) + 2��2cos(��) sin(��) a) Recall that ������2(��) + ������2(��) = 1

▪ = 2��(������2(��) + ������2(��)) →

Applications of Gradients: 2

- Directional Derivatives

����

���� = ����

▪ ���� ��(��, ��) is the directional derivative ▪ ���� ��(��) = ���� ��(��, ��)

▪ lim ��→0

��(��+��ℎ,��+ℎ��) ��(��,��)

��, where (ℎ, ��) are small differences

▪ The idea is to use the chain rule for paths when the path is a line segment joining (�� + ℎ, �� + ��) to (��, ��)

▪ ���� ��(��) = ���� ∙ ��̂, where ��̂ is the unit vector

▪ ��⃑(��) = 〈�� + ��ℎ̂, �� + ����̂〉 → ��⃑ = 〈ℎ, ��〉 = ��̂

▪ In three variables: [��(��+��ℎ,��+����,��+����)−��(��,��,��]

��

��→0∇�� ∙ ��⃑′ = ��̂ ∙ ∇��, where ��̂ = 〈ℎ, ��, ��〉

o lim

Example 3

- Suppose temperature �� = (��, ��, ��) at a point is given in degrees by: �� = 50 + ������. ▪ Find the rate of change of the temperature with respect to distance at (3,4,1) in the direction of a) ��⃑ = 〈1,2,2〉 and b) ��⃑ in the direction of the origin. a) ���� �� = ∇�� ∙ ��̂; ��̂ =��⃑⃑‖��⃑⃑‖=〈1,2,2〉

√12+22+22 =〈1,2,2〉

√9=〈1,2,2〉

3

o ∇�� = 〈����, ����, ����〉: ���� = ����; ���� = ����; ���� = ����

o ∇��(3,4,1) = 〈4,3,12〉;���� ��(3,4,1) = 〈4,3,12〉 ∙〈1,2,2〉

3=4+6+24

3

∙ ���� ��(��, ��, ��) =������

b) ��̂ =−〈3,4,1〉

√32+42+12 =−1

√26〈3,4,1〉 → ���� ��(��, ��, ��) = 〈��, ��, ����〉 ∙−��

√����〈��, ��, ��〉

▪ Find the maximum directional derivative ���� �� at the point (3,4,1) and the direction ��⃑ in which the maximum occurs.

o To maximize the directional derivative:

∙ ���� �� = ∇�� ∙ ��̂ = ‖∇��‖‖��̂‖cos(��)

∙ �� is the angle between ∇�� and ��̂ (�� = 0 → cos(��) = 1)

o The max. value of ���� �� = ‖∇��‖, when ��̂ has the same direction as ∇��. o Thus the min. occurs where �� = �� (the direction opposite of ∇��)

iClicker

- Find the slope of �� = ��(��, ��) = ��2 + ��2at (1,2) in the direction of 〈2,1〉. ▪ ∇��(1,2) = 〈2,4〉; 〈2,4〉 ∙〈2,1〉

√22+12 =4+4

��

√��

√5→

Implicit Gradients

- For any curve ��⃑(��) on a surface ��(��(��), ��(��), ��(��)) = ��.

- Hence, ��������(��(��), ��(��), ��(��)) = 0.

- ∇�� ∙ ��⃑′(��) = 0; ∇�� = 〈����, ����, ����〉

- ∇�� is the direction of the normal vector ��⃑⃑ to the surface

▪ This is true for any curve.

Example 4

- ��2 + ��2 + ��2 = 3 at the point (1, −1,1). Find the tangent plane.

▪ �� = ��2 + ��2 + ��2; ∇�� = 〈2��, 2��, 2��〉

▪ ∇��(1, −1,1) = 〈2, −2,2〉 = ��⃑⃑

o The equation is: ��⃑⃑ ∙ (��⃑ − ��⃑0) = 0

o 2(�� − 1) − 2(�� + 1) + 2(�� − 1) = 0

o �� − 1 − �� − 1 + �� − 1 →�� − �� + �� = ��

The Chain Rule (Case 2)

- Suppose �� = ��(��, ��) is a differentiable function of �� and ��, where �� = ��(��,��) and �� = ℎ(��,��) are differentiable functions of �� and ��.

- Then: ����

���� =���� ����

����

���� +���� ����

����

���� and ����

���� =����

����

����

���� +���� ����

���� ����

Applications of Gradients: 3

- Polar Coordinates

▪ �� = ��������(��), �� = ��������(��)

▪ Suppose �� is a function �� = ��(��, ��) represented as a function of (��,��) ▪����

���� =������ (��������(��)) = cos(��) ;����

▪����

���� =����

���� +���� ����

����

���� =������ (��������(��)) = sin(��)

���� = ����(−��������(��)) + ����(��������(��))

����

����

▪����

���� = ����(cos(��)) + ����(sin(��)) ���� )2+ (1������

▪ (����

����)2= ����2������2(��) + 2�������� cos(��) sin(��) + ����2������2(��) +

����2������2(��) − 2�������� cos(��) sin(��) + ����2������2(��) = ����2 + ����2 = ‖∇��‖2 ▪ Solving for ����, ���� in terms of ����, ����:

o ���� = cos(��) ���� −1��sin(��) ����

o ���� = sin(��) ���� +1��cos(��) ����

Optimization

- When the tangent plane to surface �� = ��(��, ��) is horizontal (flat), there’s a chance that the graph has a local extreme value at a point.

- However, it may be neither a maximum nor a minimum, but a saddle point. - When ∇�� = (0,0) at a point, this point is a critical point.

Example 1

- ��(��, ��) = 2��2 + ��2 − ���� − 7��. Find the local extreme.

▪ ∇�� = (4�� − ��)��̂+ (2�� − �� − 7)��̂= 〈0,0〉

▪ Solve (system of two equations):

o 4�� − �� = 0

o 2�� − �� − 7 = 0

o 2(4�� − ��) + 2�� − �� − 7 = 0

o 8�� − 2�� + 2�� − �� − 7 = 0 → 7�� = 7 → �� = ��

o Subbing back in: 4(1) − �� = 0 → �� = ��

▪ Compare values of �� at (1,4) with those at nearby points (1 + ℎ, 4 + ��): o ��(1,4) = 2 + 16 − 4 − 28 = −14

o ��(1 + ℎ, 4 + ��) = 2(1 + ℎ)2 + (4 + ��)2 − (1 + ℎ)(4 + ��) − 7(4 + ��) o ��(1 + ℎ, 4 + ��) − ��(1,4) = 2ℎ2 + ��2 − ℎ�� = (�� −ℎ2)2+74ℎ2 ≥ 0

▪ Thus, ��(��, ��) is a minimum.

Example 2

- ��(��, ��) = ��2 − ���� + 2�� + �� + 1. Find the local extreme.

▪ ∇�� = (2 − ��)��̂+ (2�� − �� + 1)��̂= 〈0,0〉

o �� = 2, �� = 5

▪ ��(5,2) = 4 − 10 + 10 + 2 + 1 = 7

▪ ��(5 + ℎ, �� + 2) = 4 + 4�� + ��2 − ℎ�� − 2ℎ − 5�� − 10 + 10 + 2ℎ + �� + 2 + 1 ▪ Subtracting the two equations: ��2 − ℎ�� + 7 − 7 → ℎ = ��

▪ Thus, ��(��, ��) is a saddle point.

Theorem

- If ��(��, ��) has continuous 2nd partial derivatives in a notion of a critical point (��, ��) and if the number �� (the discriminant) is defined by:

▪ �� = ������(��, ��) ∙ ������(��, ��) − [������(��, ��)]2, then (��, ��) is a:

a) Maximum if �� > 0 and ������(��, ��) < 0

b) Minimum if �� > 0 and ������(��, ��) > 0

c) Saddle point if �� < 0

Example 3

- �� = 2��2 + ��2 − ���� − 7��. Find the local extreme.

▪ ���� = 4�� − ��; ���� = 2�� − �� − 7

▪ ������ = 4; ������ = −1; ������ = 2

▪ �� = 4 ∙ 2 − (−1)2 = 7 → minimum

Example 4

- �� = ��2 − ���� + �� + 1

▪ ���� = −��; ���� = 2�� − �� + 1

▪ ������ = 0; ������ = −1; ������ = 2

▪ �� = 0 ∙ 2 − (−1)2 = −1 → saddle point

iClicker

- ��(��, ��) = ��2 − 2�� + ��2 − 1. Find ∇�� = 〈0,0〉.

▪ ∇�� = (2�� − 2)��̂+ (2��)��̂→(��, ��)

Important Note

- If �� = 0, no conclusion can be immediately drawn

▪ �� = 1 + 2���� − ��2 − ��2

▪ ���� = 2�� − 2��; ���� = 2�� − 2�� → �� = ��, no critical point (ridge)

- Three cases:

▪ �� = ��4 + ��4,�� = 0; (0,0) is a local minimum

▪ �� = −(��4 + ��4),�� = 0; (0,0) is a local maximum

▪ �� = ��4 − ��4,�� = 0; (0,0) is a saddle point

End of Document