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homework 9 Solutions University of California, Los Angeles CHICANO 30B

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Exercise 7.3.1 Why does the graph of a linear function have to pass through the origin? �(� + �) = �(�) + �(�) �(��) = ��(�), A linear function by definition needs to satisfy the following two conditions: Where � and � are vectors with the same number of elements a...

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  • March 6, 2023
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HW#9
Thursday, March 15, 2018 10:08 AM



Exercise 7.3.1 Why does the graph of a linear function have to pass through the origin?
Ans:
A linear function by definition needs to satisfy the following two conditions:
𝑓(𝑣 + 𝑤) = 𝑓(𝑣) + 𝑓(𝑤)
𝑓(𝑐𝑣) = 𝑐𝑓(𝑣),
Where 𝑣 and 𝑤 are vectors with the same number of elements and 𝑐 is a scalar.
Thus, a linear function must be a linear combination of some number of independent variables:
𝑓(𝑥1, 𝑥2 … 𝑥𝑘 ) = 𝑐1𝑥1 + 𝑐2𝑥2 + ⋯ + 𝑐𝑘𝑥𝑘
If (𝑥1, 𝑥2 … 𝑥𝑘 ) = (0,0, … 0) (the origin), then 𝑓(𝑥1, 𝑥2 … 𝑥𝑘 ) = 0 and the point is (0,0, … 0).

Exercise 7.3.2 In the right panel of Figure 7.6:




a) Where are the cutting planes?
Ans: The cutting planes are the grey planes 𝑋 = 0 (formed by the intersections of the 𝑌- and 𝑍-axes)
and 𝑌 = 0 (formed by the intersection of the 𝑋- and 𝑍-axes).

b) What is the significance of the red triangle?
Ans: The red triangle is a right triangle with the two legs with length Δ𝑌 and ∆𝑍. The hypotenuse is the
intersection of the green plane with the 𝑋 = 0 plane and has the slope ∆𝑍/∆𝑌.

c) How do we know that the two red lines have the same slope?
Ans: The two red lines have the same slope since they are both at the intersection of the green plane
and a cutting plane 𝑋 = constant. In other words, they are parallel to each other.

d) How do we know the values of ΔY and ΔZ (other than reading the labels)?
Ans: We know the values of Δ𝑌 and ∆𝑍 by counting the number of dashed lines (units). In order to
make the graph in 7.6, we picked two points, (0,-2,-2) and (0,2,2), at the intersection of the green plane
(𝑍 = 0.5𝑋 + 𝑌) and the 𝑋 = 0 cutting plane. Then we can calculate that ∆𝑌 = 2 − (−2) = 4 and Δ𝑍 = 2
− (−2) = 4.

Exercise 7.3.3 Compute the slope of the blue line.

Ans: Slope = Δz/Δx
= −2/4
= −1/2

Section 7.3 Further Exercises

1. Write the equation for the plane passing through the origin that has the slopes below:
a) ΔZ/ΔY = 3 and ΔZ/ΔX = 5
b) ΔZ/ΔX = 4 and ΔZ/ΔY = 1.5
c) ΔZ/ΔX = −3 and ΔZ/ΔY = −1


LS30B Page 1

, Ans: In general, a plane passing through point (𝑋0,𝑌0,, 𝑍0,) with slopes ΔZ/ΔX and ΔZ/ΔY
is given by:
∆ ∆
𝑍 − 𝑍 =⎯⎯ (𝑋 − 𝑋0) + ⎯⎯(𝑌 − 𝑌0)
∆ ∆
a) 𝑍 = 5𝑋 + 3𝑌
b) 𝑍 = 4𝑋 + 1.5𝑌
c) 𝑍 = −3𝑋 − 𝑌

2. Find ΔZ/ΔX and ΔZ/ΔY for the planes specified by the equations below:
a) Z = 7X + 25Y
b) Z = 3Y − 2X
c) Z = πY + 16X

Ans:
∆ ∆
a) ⎯⎯

= 7 ⎯⎯

= 25

∆ ∆
b) ⎯⎯ = -2 ⎯⎯ = 3
∆ ∆

∆ ∆
c) ⎯⎯ = 16 ⎯⎯ = π
∆ ∆

Exercise 7.4.1 In SageMath, plot a function of two variables. Pick a point on the function
and zoom in on it. What do you observe?

Exercise 7.4.2 Give an example of an XZ cutting plane.
Ans: An XZ cutting plane is simply any plane that has the following expression: Y = constant. In other
words, it is a plane that is parallel to the xz-coordinates plane at a particular y-valueAn 𝑋𝑍 cutting plane
has a constant value of 𝑌 (e.g. 𝑌 = 1).

Exercise 7.4.4 Find the linear approximation to Z = 4.5 − Y2/4 at Y = 3.
Ans: To find the linear approximation, we need two items:
1) The tangent point where the line touches the graph of the function.
2) The slope of the line- calculated using the point found in [1]

[1] Tangent point:

Y=3
3 18 9 9
𝑍 = 4.5 − ⎯⎯ = ⎯⎯⎯− ⎯⎯= ⎯⎯
4 4 4 4
Point = 3, ⎯

[2] Slope of the line"

d𝑧 d 𝑦
⎯⎯⎯= ⎯⎯⎯ 4.3 − ⎯⎯⎯
d𝑦 d𝑦 4
2𝑦
= 0 − ⎯⎯⎯
4
−1
= ⎯⎯⎯𝑦
2
Hence, at y =3
d𝑧 3
⎯⎯⎯= − ⎯⎯
d𝑦 2



LS30B Page 2

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