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# C3 question Watch

1. So I'm being an idiot clearly but can't see what I'm doing wrong.

Question 7 part 2. dx/dy is .

So then I said that at A dx/dy =0 but then i get stuck.
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2. (Original post by Music99)
So I'm being an idiot clearly but can't see what I'm doing wrong.

Question 7 part 2. dx/dy is .

So then I said that at A dx/dy =0 but then i get stuck.
I think you're thinking A is a turning point but the question doesn't say that.

You can find the gradient of the curve at a point by substituting the coordinate into dy/dx. At A, y=0. I'll let you work out the y-coordinate of B.
3. (Original post by notnek)
I think you're thinking A is a turning point but the question doesn't say that.

You can find the gradient of the curve at a point by substituting the coordinate into dy/dx. At A, y=0. I'll let you work out the y-coordinate of B.
Ohhhh! Facepalm!!
4. (Original post by notnek)
I think you're thinking A is a turning point but the question doesn't say that.

You can find the gradient of the curve at a point by substituting the coordinate into dy/dx. At A, y=0. I'll let you work out the y-coordinate of B.
So I subbed y=0 into dx/dy and you get ln(3) +8/3 but that doesn't give you the correct gradient.
5. (Original post by Music99)
So I subbed y=0 into dx/dy and you get ln(3) +8/3 but that doesn't give you the correct gradient.
dy/dx gives you the gradient, not dx/dy..
6. (Original post by Music99)
So I'm being an idiot clearly but can't see what I'm doing wrong.

Question 7 part 2. dx/dy is .

So then I said that at A dx/dy =0 but then i get stuck.
can you send me that paper please?
7. (Original post by notnek)
dy/dx gives you the gradient, not dx/dy..
I thought that because at A y=0 we would use dx/dy and at B use dy/dx ...
8. (Original post by Music99)
I thought that because at A y=0 we would use dx/dy and at B use dy/dx ...
The gradient is always dy/dx (if the y-axis is vertical). Substituting the x-coordinate into dy/dx (x) or the y-coordinate into dy/dx (y) will give you the same answer.
9. (Original post by notnek)
The gradient is always dy/dx (if the y-axis is vertical). Substituting the x-coordinate into dy/dx (x) or the y-coordinate into dy/dx (y) will give you the same answer.
Ah okay, so then other then finding the rate at which x changes w.r.t y is there a use for dx/dy
10. (Original post by Music99)
Ah okay, so then other then finding the rate at which x changes w.r.t y is there a use for dx/dy
That sounds like a good use to me. What use that dy/dx has would you like dx/dy to have?

11. (Original post by notnek)
That sounds like a good use to me. What use that dy/dx has would you like dx/dy to have?

Haha, :P, going to mark the paper soon, If I get things wrong mind If I quote you?
12. (Original post by Music99)
Haha, :P, going to mark the paper soon, If I get things wrong mind If I quote you?
Sure.

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