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Morgarella [4.7K]
3 years ago
12

A normal line is a line that is perpendicular to another line, curve or surface. Give an everyday example of a normal line to ea

ch of the following: another line, a circle, a plane.

Mathematics
2 answers:
Rudiy273 years ago
8 0

Answer:

a) A cross b) A stool c)  One foot table

Step-by-step explanation:

Hi there!

Check the pictures below for better visualizing.

a) Cross

The figure of a cross shows clearly one normal line perpendicular to another perpendicular one. They cross each other.

b) Circle

A stool has a normal line perpendicular to the ground.

c) One foot table

In this Table, we have a normal line perpendicular to the ground and its top.

tester [92]3 years ago
4 0

Here are the examples for

1. a normal line to a line

The corner of a photo frame, or the corner of a note book

2. A normal line to a circle

A circular plate balanced on a scale

3. A normal line to a plane

The pole standing on a plain ground

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For the function defined by f(t)=2-t, 0≤t&lt;1, sketch 3 periods and find:
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The half-range sine series is the expansion for f(t) with the assumption that f(t) is considered to be an odd function over its full range, -1. So for (a), you're essentially finding the full range expansion of the function

f(t)=\begin{cases}2-t&\text{for }0\le t

with period 2 so that f(t)=f(t+2n) for |t| and integers n.

Now, since f(t) is odd, there is no cosine series (you find the cosine series coefficients would vanish), leaving you with

f(t)=\displaystyle\sum_{n\ge1}b_n\sin\frac{n\pi t}L

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b_n=\displaystyle\frac2L\int_0^Lf(t)\sin\frac{n\pi t}L\,\mathrm dt

In this case, L=1, so

b_n=\displaystyle2\int_0^1(2-t)\sin n\pi t\,\mathrm dt
b_n=\dfrac4{n\pi}-\dfrac{2\cos n\pi}{n\pi}-\dfrac{2\sin n\pi}{n^2\pi^2}
b_n=\dfrac{4-2(-1)^n}{n\pi}

The half-range sine series expansion for f(t) is then

f(t)\sim\displaystyle\sum_{n\ge1}\frac{4-2(-1)^n}{n\pi}\sin n\pi t

which can be further simplified by considering the even/odd cases of n, but there's no need for that here.

The half-range cosine series is computed similarly, this time assuming f(t) is even/symmetric across its full range. In other words, you are finding the full range series expansion for

f(t)=\begin{cases}2-t&\text{for }0\le t

Now the sine series expansion vanishes, leaving you with

f(t)\sim\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos\frac{n\pi t}L

where

a_n=\displaystyle\frac2L\int_0^Lf(t)\cos\frac{n\pi t}L\,\mathrm dt

for n\ge0. Again, L=1. You should find that

a_0=\displaystyle2\int_0^1(2-t)\,\mathrm dt=3

a_n=\displaystyle2\int_0^1(2-t)\cos n\pi t\,\mathrm dt
a_n=\dfrac2{n^2\pi^2}-\dfrac{2\cos n\pi}{n^2\pi^2}+\dfrac{2\sin n\pi}{n\pi}
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Here, splitting into even/odd cases actually reduces this further. Notice that when n is even, the expression above simplifies to

a_{n=2k}=\dfrac{2-2(-1)^{2k}}{(2k)^2\pi^2}=0

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f(t)\sim\dfrac32+\displaystyle\sum_{n\ge1}a_n\cos n\pi t
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f(t)\sim\dfrac32+\displaystyle\sum_{k\ge1}\frac4{(2k-1)^2\pi^2}\cos(2k-1)\pi t

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