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**Zeeshan 01****Member**- Registered: 2016-07-22
- Posts: 648

Help for this question

Find y2

X^2+y^2=a^2

MZk

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**CIV****Member**- Registered: 2014-11-09
- Posts: 69

You first. What have you done to try and solve this? Show us.

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**Zeeshan 01****Member**- Registered: 2016-07-22
- Posts: 648

I solve dy by dx how further solve for y2

MZk

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**CIV****Member**- Registered: 2014-11-09
- Posts: 69

Someone else will have to show you, because I don't see an effort being made here. Just a word of advice... if you keep posting like this, eventually people will stop wanting to help you. You make us feel like we're doing your homework for you. That of course is just my opinion.

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**CIV****Member**- Registered: 2014-11-09
- Posts: 69

This is a question I had asked: http://www.mathisfunforum.com/viewtopic.php?id=23783

I made a serious effort to do the problem and made sure that everyone could see this. It is homework for me, but clearly you can see that I'm trying.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,168

Zeeshan 01 wrote:

Find y2

X^2+y^2=a^2

I do not understand what this question is asking. y2 means what?

If this is a differentiation question, please post what you got for dy/dx.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Zeeshan 01****Member**- Registered: 2016-07-22
- Posts: 648

It's second derivative

MZk

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,168

hi Zeeshan 01

Thanks for clearing that up. You can write it like this in LaTex:

As your expression contains y^2 it is possible you haven't got dy/dx correct yet. Expressions like that are more difficult. There's no point trying to show you the next step until I know you have that first step correct. So please do as asked and post your working so far. Thanks,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Zeeshan 01****Member**- Registered: 2016-07-22
- Posts: 648

When we take dy/dx it's not equal to zero when we take da/dx it's equal to zero how ????

MZk

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,168

hi Zeeshan 01

in post 3 you wrote:

I solve dy by dx

Now you are asking a new question? Third time of asking ... Please post your answer to dy/dx.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Zeeshan 01****Member**- Registered: 2016-07-22
- Posts: 648

Minus y divide by x is ans

MZk

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,168

hi Zeeshan 01

Excellent! That tells me two things. (1) You know about 'implicit' differentiation (2) That's the way you'll want to do the next step.

But first let's just recap how you got there.

So x and y are variables and 'a' is a constant.

When we differentiate that equation each part must be differentiated with respect to (wrt) x.

x^2 is easy enough. That becomes 2x.

But what about y^2. This is not even a function of x. You have to make use of the chain rule:

In this case z = y^2

So we differentiate wrt y and then multiply by dy/dx. So we get

Now we must differentiate the a^2. But 'a' is a constant. It's gradient function is zero whatever x is. So

Putting all this together and re-arranging you end up with

Now to differentiate this.

The right hand side is a 'quotient' or fraction so we must apply the rule for that:

If you have not met this rule before, please ask about it.

In your example p = y and q = x so we get

You can do a little simplifying from here.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**Zeeshan 01****Member**- Registered: 2016-07-22
- Posts: 648

If you have not met this rule before, please ask about it.

Hehe I am well aware

*Last edited by Zeeshan 01 (2017-03-15 03:56:02)*

MZk

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**Zeeshan 01****Member**- Registered: 2016-07-22
- Posts: 648

Why we differentate wrt x???

MZk

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**Zeeshan 01****Member**- Registered: 2016-07-22
- Posts: 648

Dy /dx is not equal to zero but da\dx is zro plz explain? ??

MZk

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**Zeeshan 01****Member**- Registered: 2016-07-22
- Posts: 648

Now we must differentiate the a^2. But 'a' is a constant. It's gradient function is zero whatever x is.

What is gradient function!!!!

MZk

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,168

hi Zeeshan 01

When you differentiate y wrt x you are finding the gradient function of y . That is what differentiation is.

have a look at these two graphs.

In the first, pick any point on the graph and imagine a line drawn at that point with the same gradient as the curve. We say the line is a tangent to the curve at that point. Make a note of the gradient at that x coordinate and consider on a new graph the point (x, gradient). Do that for all of the points on the curve and you get a new curve which is called the gradient function of y.

For y = x^2 let's think about what this gradient function is like. when x is positive, the gradient is positive. Because the curve y = x^2 is symmetrical the gradient at a negative value of x will be the same as at the positive value of x, except it will now be negative. For example, at x = 2 there will be a certain gradient. At x = -2 the line slopes exactly the same amount but downwards from left to right. So if the gradient at x = 2 is m, then the gradient at x = -2 will be -m. At x = 0 we can see that the gradient is also 0.

It is possible to show that the gradient function is 2x so we can write dy/dx = 2x. But even without knowing the theory for this you should be able to see that 2x has the right properties. It is zero at x=0; at x = 3 it is 6; and at x = -3 it is -6.

Now what about the gradient function for a^2. The graph plotter required that I choose a value for a, so I chose a = 3. The graph for y = a^2 is a straight horizontal line. The gradient is zero at every point. So its gradient function is zero at every point. So d(a^2)/dx = 0.

If 'a' varies as x varies then this would not be so. But, in the context of your question, it seemed reasonable to assume that a is a constant. If that is not so then d(a^2)/dx = 2a.(da/dx). In other words, it would have a similar form to the gradient function for y^2.

Why differentiate wrt x? Because you wanted dy/dx. The dx tells us what you have to differentiate with respect to.

Hope that explanation helps. So have you got a final answer for y2 ?

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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