Question

can someone one check my work ... :)

Answer 1

@SolomonZelman

Answer 2

actually I got y^2-4.

Answer 3

i personally dont see any mistakes

Answer 4

oh hmmm

Answer 5

wait ...

Answer 6

You are right ...

\(\color{black}{\displaystyle (2-y)^2+4(2-y) }\)
\(\color{black}{\displaystyle ((2-y)+4)(2-y) }\)
\(\color{black}{\displaystyle (6-y)(2-y) }\)
\(\color{black}{\displaystyle (y-2)(y-6) }\)
\(\color{black}{\displaystyle y^2-8y+12 }\)

Answer 7

I solved t=y-2 instead.
(I'm off today for some reason (just like Timberwolves))

Answer 8

oh okay x)

Answer 9

you are right!

Answer 10

yays xD

hmm can you check this one too... not sure what to do next
one sec

Answer 11

same directions from the previous picture

Answer 12

\(\color{black}{\displaystyle \sec \theta =\frac{1}{\cos \theta}\quad \Longleftrightarrow \quad y=\frac{1}{x}.}\)

The parameter is defined, \(\color{black}{\displaystyle t\in [0,\pi/2] }\), and since the function is monotonic (monotonically decreases) over this interval we can safely say that the domain is going to be bound by \(\color{black}{\displaystyle x\in [\cos(\pi/2),\cos(0)] }\).

Answer 13

oh so how did you get y= 1/x?

Answer 14

\(\color{black}{\displaystyle \sec \theta =\frac{1}{\cos \theta}}\) Yes?

Answer 15

yeh

Answer 16

\(\color{black}{\displaystyle y=\sec \theta }\) and \(\color{black}{\displaystyle x=\cos \theta}\).
Right?

Answer 17

These are the given parametric expressions of your variables.

Answer 18

yeah

Answer 19

So, we just substitute:)

Answer 20

\(\color{black}{\displaystyle \sec \theta =\frac{1}{\cos \theta}}\)


\(\color{black}{\displaystyle {\rm Given:{~}}xy= \sec \theta {\rm{~~~}and{~~~}}x=\cos \theta}\)

\(\color{black}{\displaystyle y =\frac{1}{x}{\rm{~~~}via{~}substitution.}}\)

Answer 21

Clear on the y=1/x :)

Answer 22

oh yeah.. i see it lol i was setting them equal to each other for some reason

Answer 23

okay so then how would the graph look like ?

do i graph the y=1/x or the two given functions

Answer 24

You are graphing y=1/x, and the domain is to be defined ... I said it, technically, but I didn't have time thus far to make a note on it.

Answer 25

oh okay so when doing these type of problems.. do we usually get the new equations and not graph the two functions that we are given ?

Answer 26

Yes, you just need to eliminate the parameter (to get some expression in x and y), and then to find the domain (or range) that corresponds to the domain of the parameter.

Answer 27

Your function (y=1/x) is monotonically decreasing. Thus, you know
that your endpoints are max and min. More specifically, we infer that
the first value of t (t=0) is the upper bound, and the second value of
t (t=π/2) is the lower bound.

We had: \(\color{black}{\displaystyle t\in [0,\pi/2] }\).

and noted the above, we have:
\(\color{black}{\displaystyle x\in [\cos(\pi/2),\cos(0)] }\), or
\(\color{black}{\displaystyle x\in [0,1] }\).

Answer 28

If you look back at the parametric function, it was undefined at t=π/2,
(so really the correct domain is \(\color{black}{\displaystyle t\in [0,\pi/2) }\) ...)

and, this lack of value at \(\pi/2\) is consistent with the lack of value at \(\color{black}{\displaystyle x=0 }\)
(x=0 corresponds to t=pi/2, as I explained before)

Answer 29

You can fix it after eliminating the parameter.
That is -
we obtain \(\color{black}{\displaystyle x\in [0,1] }\), but obviously the function y=1/x is undefined at x=0, so simply we will fix this to \(\color{black}{\displaystyle x\in (0,1] }\).

OR, you can fix it at the beginning.
That is -
At \(\color{black}{\displaystyle t=\pi/2}\), \(\color{black}{\displaystyle y=\sec t}\) is undefined , so we will adjust to this, by saying, \(\color{black}{\displaystyle t\in [0,\pi/2)}\).

Answer 30

Either way, you end up getting the same domain for x, and excuse me for using t instead of \(\theta\).

Answer 31

oh okay i see now.. it makes sense

Answer 32

Also, you can use parametric calculus to maximize and minimize the function, in case where the function doesn't behave monotonically over some domain for t.

Answer 33

Alrighty:)

Answer 34

thank you :D

Answer 35

No Problem:)