Question

The largest value of a function

Answer 1

How to find the largest value of a function\[q(1-\cos\ \theta)+\sin\theta\] in the range \[0 \le \theta \le \alpha + \pi\]

Answer 2

Value of α is given.

Answer 3

you know the highest value for cos and sin?

Answer 4

any information on q ? is it positive and greater than 1

Answer 5

q is positive, but I don't know if it's greater that 1.
\[q=\frac{ GM }{ pv^2 }\] where G is universal gravitational constant, M is mass of the Sun, p is distance from the Sun to the body and v is velocity of the body. Values for p and v are not given.

Answer 6

I guess theta is the variable here then differentiate WRT to theta and make the result equal to zero

Answer 7

what is WRT?

Answer 8

with respect to

Answer 9

\[q \sin\ \theta+\cos\theta=0\]
this?

Answer 10

yes now you know the value of theta where function gets maximized

Answer 11

Ok, Thanks. But in the book it's written this:
"The largest value attained by this function in the range \[0 \le \theta \le \alpha+\pi\] is \[q+(q^2+1)^{1/2}\]"

Answer 12

can you explain me this?

Answer 13

@gorica not sure but let me try I will message if I figure this out

Answer 14

ok, thanks. If it's easier for you, I can send you that page of the book?

Answer 15

or the whole problem and solution

Answer 16

k

Answer 17

I also don't understand when they calculate deflection angle, they first say α=π-θ, which is ok, but when they calculate tg(α/2), they wrote tg(α/2)=tg(θ/2-π/2)

Answer 18

I think I figured it out! we have already calculated the value of tan(theta) by differentiating. Now you have to use the formula 1+tan^2 = sec^

Answer 19

in the original equation divide all the terms by cos and proceed it should give you the answer

Answer 20

ok, I will try now :)

Answer 21

Hope this helps! I have not done the complete calculation. let me know if you get stuck

Answer 22

original equation you mean q(1-cos(theta))+sin(theta)?

Answer 23

yeah one more thing to take care...tan(theta) was negative so we know theta will lie in second quadrant hence cos(theta) will also be negative
Difficult Question this is!

Answer 24

so theta actually is not from [0, alpha+pi] as it's said, but from [pi/2,pi]? I think this could be concluded looking at the equation. There is - before cos(theta), so it will be greater value if cosinus is negative and sinus is positive.
Anyway, I didn't get required by dividing by cos(theta) :/

Answer 25

No need for that can you find value of sin and cos from the tan?

Answer 26

you can use 1+tan^2= sec^2 and 1+cot^2=csc^2 to find sin and cos then substitute these value in the equation

Answer 27

@gorica can you solve from here?

Answer 28

I will try :)

Answer 29

no, I can't get it :/

Answer 30

what are values of sin and cos in terms of q?

Answer 31

From identities that you gave me, I've got that\[\cos^2\theta=\frac{ 1 }{ 1+q^2 }\]\[\sin^2\theta=\frac{ q^2 }{ q^2+1 }\]

Answer 32

are you sure? I think you have interchanged value of cos and sin :)

Answer 33

yes, sorry. I wrote that q=tg(theta) :) ok, I think I have to try again :D

Answer 34

tg(theta)=-1/q

Answer 35

I will start to write it here because I'm just going in the circle.

Answer 36

\[q(1-\cos \theta)+\sin \theta=q(1+\sqrt{(\frac{ q^2 }{ q^2+1 })})+\sqrt{\frac{ 1 }{ q^2+1 }}\]
\[=q+q \sqrt {q^2}\sqrt{\frac{ 1 }{ q^2+1 }}+\sqrt{\frac{ 1 }{ q^2+1 }}\]\[=q+q^2 \sqrt{\frac{ 1 }{ q^2+1 }}+\sqrt{\frac{ 1 }{ q^2+1 }}\]\[=q+\sqrt{\frac{ 1 }{ q^2+1 }}(q^2+1)\]\[=q+(q^2+1)^{-1/2}(q^2+1)\]\[=q+(q^2+1)^{1/2}\]

Answer 37

:)

Answer 38

Great work!

Answer 39

Thank you very much! :)

Answer 40

No problem! just for the information which grade are you in?