Turnig points

Anthonyk2013

Junior Member
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Sep 15, 2013
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X=Ѳ(6-Ѳ)

I have to find the turning points of this problem. I have completed the first 6 questions but this is confusing me.

In the other questions I find dy/dx then I can find X and find Y, here I'm not sure where to start.

Can anyone point me in the right direction.
 
X=Ѳ(6-Ѳ)

I have to find the turning points of this problem. I have completed the first 6 questions but this is confusing me.

In the other questions I find dy/dx then I can find X and find Y, here I'm not sure where to start.

Can anyone point me in the right direction.

Since X and Ѳ are what are called dummy variables, what would you do if you replaced X by y and Ѳ by x for
y = x ( 6 - x) = 6x - x2?
 
What if you'd been given a function like x=3y^2+5y+7? Could you have found the turning points of that? As a hint, remember that you can find the derivative with respect to a variable other than x.
 
What if you'd been given a function like x=3y^2+5y+7? Could you have found the turning points of that? As a hint, remember that you can find the derivative with respect to a variable other than x.

Id have no problem with x=3y^2+5y+7. X=Ѳ(6-Ѳ) is confusing me a bit.
 
That's very close to the answer. Ishuda gave you good advice, except for one point on which I disagree. In this problem, theta is indeed a "dummy" variable" and can be replaced by y without causing any problems. However, x is not a dummy variable in this problem. Consider the function x=y(6-y). That forms a parabola which opens to the left. Now consider the function y=x(6-x). That forms a parabola which opens down. So, you can see that the turning points of these two functions will not be the same. Here, it definitely matters that x is a function of some other variable.

With your answer, you're saying that the turning point of the function is when x=3 and theta (y) = 9. But that can't possibly be a turning point for the function, because that point isn't on the parabola. What do you think you need to do to get the correct answer?
 
That's very close to the answer. Ishuda gave you good advice, except for one point on which I disagree. In this problem, theta is indeed a "dummy" variable" and can be replaced by y without causing any problems. However, x is not a dummy variable in this problem. Consider the function x=y(6-y). That forms a parabola which opens to the left. Now consider the function y=x(6-x). That forms a parabola which opens down. So, you can see that the turning points of these two functions will not be the same. Here, it definitely matters that x is a function of some other variable.

With your answer, you're saying that the turning point of the function is when x=3 and theta (y) = 9. But that can't possibly be a turning point for the function, because that point isn't on the parabola. What do you think you need to do to get the correct answer?

I'm not a 100% sure what to do next if I'm honest, Ѳ normally represents an angle?
 
Yes, theta normally represents an angle, but it doesn't always have to. Theta is just a Greek letter which we are using as a generic variable. This is a rather silly example, but theta could represent the number of bread crumbs you throw into the local pond, and then x as a function of theta represents how many ducks land in the pond to eat the bread.

Also, I see Ishuda has posted again. I see now that my earlier post was not 100% right. As noted, if you plot the coordinates in the form (theta, x) then (3,9) is a valid turning point. In my post, I opted for the traditional coordinate system that I learned where x​ always comes first.
 
Yes, theta normally represents an angle, but it doesn't always have to. Theta is just a Greek letter which we are using as a generic variable. This is a rather silly example, but theta could represent the number of bread crumbs you throw into the local pond, and then x as a function of theta represents how many ducks land in the pond to eat the bread.

Also, I see Ishuda has posted again. I see now that my earlier post was not 100% right. As noted, if you plot the coordinates in the form (theta, x) then (3,9) is a valid turning point. In my post, I opted for the traditional coordinate system that I learned where x​ always comes first.

Ok thanks
 
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