FP1 - Edexcel - Coordinate systems - Need help with question

I had this exact same conversation with another user on TSR. Let me try and find that thread...

While I'm looking though, think of the graph of

y = \sqrt{x}

. You see that the range is always positive.

(edited 11 years ago)

Reply 2

OP

Original post by claret_n_blue

I had this exact same conversation with another user on TSR. Let me try and find that thread...

Thanks a lot! I thought I was being silly and missing something.

Reply 3

OP

Original post by claret_n_blue

I had this exact same conversation with another user on TSR. Let me try and find that thread...

While I'm looking though, think of the graph of

y = \sqrt{x}

. You see that the range is always positive.

I can't make the link as earlier in the book they use the positive and negative of y=...

Yet for some reason it doesn't apply in this case and I don't understand why.

Reply 4

Original post by Jackabc

Thanks a lot! I thought I was being silly and missing something.

http://www.thestudentroom.co.uk/showthread.php?t=2214694

They basically say that

(\pm 2)^2 = 4

however

\sqrt{4} = 2

.

This website explains a little more about the "principal square root".

Reply 5

OP

Original post by claret_n_blue

http://www.thestudentroom.co.uk/showthread.php?t=2214694

They basically say that

(\pm 2)^2 = 4

however

\sqrt{4} = 2

.

This website explains a little more about the "principal square root".

So why is it when x^2=9 you give x=3 or x=-3

Reply 6

Original post by Jackabc

So why is it when x^2=9 you give x=3 or x=-3

That's because either

(-3) \cdot (-3) = 9

or

3 \cdot 3 = 9

. But

\sqrt{9} = +3

.

EDIT: The square root symbol itself represents the positive value (hence the graph). But you know that there are two possible values and so we put the negative possible value in ourselves as we're not sure if we want the positive or negative solution.

What's never been made clear (when it should be) is that some

x

such that

x^2 = y

is not the same as

\sqrt{y}

.

(edited 11 years ago)

Reply 7

OP

Original post by claret_n_blue

That's because either

(-3) \cdot (-3) = 9

or

3 \cdot 3 = 9

. But

\sqrt{9} = +3

.

Can you expand on that a bit? That website isn't that clear.

So x^2=9 so x is equal to the square root of 9. Then if the square root of 9 is 3 then why bother with -3.

Reply 8

OP

Original post by claret_n_blue

That's because either

(-3) \cdot (-3) = 9

or

3 \cdot 3 = 9

. But

\sqrt{9} = +3

.

EDIT: The square root symbol itself represents the positive value (hence the graph). But you know that there are two possible values and so we put the negative possible value in ourselves as we're not sure if we want the positive or negative solution.

What's never been made clear (when it should be) is that some

x

such that

x^2 = y

is not the same as

\sqrt{y}

.

That makes sense now. So really we are not thinking the square root of 9, we are thinking what could be multiplied to give 9?

Reply 9

Original post by Jackabc

That makes sense now. So really we are not thinking the square root of 9, we are thinking what could be multiplied to give 9?

~~Yep! Spot on~~

EDIT: I re-read your post properly. Sorry, if you mean what is multiplied by itself to give you 9, that is the same as taking the square root. When you are taking a square root, you have the possibility to choose from two (the positive or negative one). The convention is to choose the positive one.

(edited 11 years ago)

Reply 10

I'd suggest implicit differentiation on that specific question (or questions similar to) to avoid that confusion.

Posted from TSR Mobile

Reply 11

OP

Original post by claret_n_blue

Yep! Spot on :smile:

As you can see in this question both the positive and negative square root are taken. Why is it in this case only the positive square root can be taken? As t could be positive or negative, t could be negative and y would be below the graph so you would be looking for the negative gradient of the tangent and the graph of the negative square root.

photo(1).jpg513.6KB

Reply 12

OP

Original post by Joshmeid

I'd suggest implicit differentiation on that specific question (or questions similar to) to avoid that confusion.

Posted from TSR Mobile

But if you look at my above example they take the negative version and how do I know with example 11, the first example that t is not negative. Would implicit differentiation be reliable then?

Reply 13

Original post by Jackabc

But if you look at my above example they take the negative version and how do I know with example 11, the first example that t is not negative. Would implicit differentiation be reliable then?

Differentiating implicitly jumps from the first line straight to the 8th line or so, much faster and avoids confusion.

We have

y^{2} = 27x \rightarrow 2y\frac{dy}{dx} = 27 \rightarrow \frac{dy}{dx} = \frac{27}{2y}

.

We can then substitute in y and find the gradient of the tangent at that specific point.

Posted from TSR Mobile

Reply 14

davros

16

Original post by Jackabc

But if you look at my above example they take the negative version and how do I know with example 11, the first example that t is not negative. Would implicit differentiation be reliable then?

YES! I have to go out now but there are "safer" methods of getting the same answer - I'll post back in a bit :smile:

Reply 15

OP

Original post by Joshmeid

Differentiating implicitly jumps from the first line straight to the 8th line or so, much faster and avoids confusion.

We have

y^{2} = 27x \rightarrow 2y\frac{dy}{dx} = 27 \rightarrow \frac{dy}{dx} = \frac{27}{2y}

.

We can then substitute in y and find the gradient of the tangent at that specific point.

Posted from TSR Mobile

I get you, thanks a lot! I will use that in future but could you tell me where using the previous method breaks down if I took the positive and negative square root.

Reply 16

OP

Original post by davros

YES! I have to go out now but there are "safer" methods of getting the same answer - I'll post back in a bit :smile:

Thanks a lot. My question now is why wouldn't the previous method work is that t could be negative and so what's is wrong with taking the negative square root of y and then working out the gradient and the normal from there.

Reply 17

Original post by Jackabc

I get you, thanks a lot! I will use that in future but could you tell me where using the previous method breaks down if I took the positive and negative square root.

Depends on the question, in the second question it refers to both points on the curve and you end up finding equations for the tangent on both, the gradient for one tangent is positive, the gradient of the other is negative as you can see from the image.

For example 11, we can see that the equation of the normal has a negative gradient(- t), therefore the gradient of the tangent at point P will be positive so we take the positive value of y.

Posted from TSR Mobile

(edited 11 years ago)

Reply 18

OP

Original post by Joshmeid

Depends on the question, in the second question it refers to both points on the curve and you end up finding equations for the tangent on both, the gradient for one tangent is positive, the gradient of the other is negative as you can see from the image.

For example 11, we can see that the equation of the normal has a negative gradient(- t), therefore the gradient of the tangent at point P will be positive so we take the positive value of y.

Posted from TSR Mobile

I think that's where I am going wrong, how do we know from example 11 that t has a negative gradient? As if t was negative then wouldn't the equation I am supposed to be proving have a positive gradient. It is weird but it seems that it seems to work whatever the case.

Reply 19