Differentiate y = x/(sqrt[sin [x]]), verify stationary point

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Monkeyseat

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Hi,

Question

Differentiate y = x/(sqrt[sin [x]]), giving your answer in its simplest form, and show that there is a stationary point between x = 2 and x = 3.

Working

I think I've differentiated and simplified it okay, so I'll save you the working. I believe it is:

dy/dx = (2sin[x] - xcos[x])/(2(sin[x])^(3/2))

However, I don't know how to show that there is a stationary point between 2 and 3.

I thought I could possibly do this:

(2sin[x] - xcos[x])/(2(sin[x])^(3/2)) = 0
2sin[x] - xcos[x] = 0
2sin[x] = xcos[x]
2(sin[x]/cos[x]) = x
2tan[x] = x
tan[x] = x/2

I then sketched tan[x] = x/2 but it didn't really help. I also subbed x = 2 and x = 3 into dy/dx but that didn't help either.

I'm not sure how to approach this. Please could you help me with the second part: "show that there is a stationary point between x = 2 and x = 3".

Many thanks.
 
Monkeyseat said:
Hi,

Question

Differentiate y = x/(sqrt[sin [x]]), giving your answer in its simplest form, and show that there is a stationary point between x = 2 and x = 3.

Working

I think I've differentiated and simplified it okay, so I'll save you the working. I believe it is:

dy/dx = (2sin[x] - xcos[x])/(2(sin[x])^(3/2))

However, I don't know how to show that there is a stationary point between 2 and 3.

I thought I could possibly do this:

(2sin[x] - xcos[x])/(2(sin[x])^(3/2)) = 0
2sin[x] - xcos[x] = 0
2sin[x] = xcos[x]
2(sin[x]/cos[x]) = x
2tan[x] = x
tan[x] = x/2

I then sketched tan[x] = x/2 but it didn't really help. I also subbed x = 2 and x = 3 into dy/dx but that didn't help either.

I'm not sure how to approach this. Please could you help me with the second part: "show that there is a stationary point between x = 2 and x = 3".

Many thanks.
Click to expand...

f(x) = x/sqrt(sin(x)) does not have a stationary point between x = 2 and x = 3.

Plot the function and you'll see.
 
\(\displaystyle y = x \cdot \sqrt{\sin{x}}\)
has a stationary point between 2 and 3.
 
I thought this was confusing. I have definitely copied it correctly though. So the question is wrong? Thanks for clarifying.

Can I just ask, if the question was to verify that there is a stationary point between x = 2 and x = 3 for y = x * (sqrt[sin[x]]) then, how would I do this?

y = x * (sqrt[sin[x]])
dy/dx = (xcos[x] + 2sin[x])/(2sqrt[sin[x]])

Okay, so there is a stationary point when dy/dx = 0.

(xcos[x] + 2sin[x])/(2sqrt[sin[x]]) = 0
xcos[x] + 2sin[x] = 0
2sin[x] = -xcos[x]
2 * (sin[x]/cos[x]) = -x
2tan[x] = -x
tan[x] = -x/2

So, to prove that there is a stationary point between 2 and 3, would it be acceptable to sketch the graphs of y = tan[x] and y = -x/2 and show that they intercept between 2 and 3? Also, is there another way/is it possible to solve this, or would the way I suggested suffice (it says 'show' that there is a stationary point but I don't know whether that means find the coordinates)?

Thanks.
 
In this case, there are no closed form solutions to the equation of interest (2tan(x)-x =0). So graphical solution is valid - there are ofcourse numerical mehods to solve for 'x'. In some cases analytical soluiton is possible (e.g. x[sup:5no1sqdd]2[/sup:5no1sqdd]+x-1 =0). However, in my book, graphical solution is always - valid unless explicitly forbidden.
 
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