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Calulating the gradient/slope of a curve

Hello there

Please can anyone explain to me how to calculate the gradient/slope of this curve between points 90s and 120s (the fastest rate of reaction)? The coordinates are (90,0.246) and (120, 0.179).

I always thought I understood how to calculate the gradient of a line, but it's recently come to my attention that the equation changes depending on whether the curve is positive (as x increases, y increases) or negative (as x increases, y decreases).

This particular one is negative, and I'm unsure what I must to calculate the slope between the 2 point 90s and 120s.

I know the formula y2-x2/x2-y2, but does that apply here (to a negative curve) and, if so, why? And which y coordinate is y2 and which x coordinate is x2?

Cell biology photosynthesis graph 2.jpg

Thank you very much, any help is greatly appreciated!
Reply 1
You have two coordinates, (x1, y1) and (x2, y2).
dy/dx = (y2 - y1)/(x2 - x1)
= -(y2 - y1)/-(x2 - x1) [times top and bottom by -1/-1]
= (y1 - y2)/(x1 - x2)

i.e. dy/dx = (y2 - y1)/(x2 - x1) = (y1 - y2)/(x1 - x2)

What does this tell us? It doesn't matter which coordinate we pick as our x1/2, y1/2 as long as we keep consistent.

Just try out the formula (your one appears to be incorrect), and you'll notice that the gradient is automatically negative.
Reply 2
Original post by Oshi
You have two coordinates, (x1, y1) and (x2, y2).
dy/dx = (y2 - y1)/(x2 - x1)
= -(y2 - y1)/-(x2 - x1) [times top and bottom by -1/-1]
= (y1 - y2)/(x1 - x2)

i.e. dy/dx = (y2 - y1)/(x2 - x1) = (y1 - y2)/(x1 - x2)

What does this tell us? It doesn't matter which coordinate we pick as our x1/2, y1/2 as long as we keep consistent.

Just try out the formula (your one appears to be incorrect), and you'll notice that the gradient is automatically negative.


Thank you very much for your reply!

Why and when would I do this stage? = -(y2 - y1)/-(x2 - x1) [times top and bottom by -1/-1]
When would it be -(y2-y1)/-(x2-x1)? Would this be if the curve is negative?

I've tried doing it both ways and see I get an answer of -2.23E-3 either way. Is this correct?
Reply 3
Original post by Ggdf
Thank you very much for your reply!

Why and when would I do this stage? = -(y2 - y1)/-(x2 - x1) [times top and bottom by -1/-1]
When would it be -(y2-y1)/-(x2-x1)? Would this be if the curve is negative?

I've tried doing it both ways and see I get an answer of -2.23E-3 either way. Is this correct?


That's the correct answer.

You don't need to do that part - I was demonstrating the reason you can choose either point for your (x1, y1) and (x2, y2). Multiplying by -1/-1 is the same thing as multiplying by 1. Don't worry too much about that, just read the summary.

Summary: it doesn't matter whether your line is positive, negative, whatever.
You've got two points, (x1, y1) and (x2, y2).
Plug them into the formula dy/dx = (y2 - y1)/(x2 - x1),
and you have your answer.
Reply 4
Original post by Oshi
That's the correct answer.

You don't need to do that part - I was demonstrating the reason you can choose either point for your (x1, y1) and (x2, y2). Multiplying by -1/-1 is the same thing as multiplying by 1. Don't worry too much about that, just read the summary.

Summary: it doesn't matter whether your line is positive, negative, whatever.
You've got two points, (x1, y1) and (x2, y2).
Plug them into the formula dy/dx = (y2 - y1)/(x2 - x1),
and you have your answer.


Thank you very much for your help. Is there a general rule that the smallest number of the 2 y coordinates should always be subtracted from the largest? And the same with the x coordinates?

Also, do you thing this looks like a suitable tangent to the curve to draw prior to calculating the gradient?

Cell biology photosynthesis graph 3.jpg

I really am grateful for your help!
Reply 5
Original post by Ggdf
Thank you very much for your help. Is there a general rule that the smallest number of the 2 y coordinates should always be subtracted from the largest? And the same with the x coordinates?

Also, do you thing this looks like a suitable tangent to the curve to draw prior to calculating the gradient?



I really am grateful for your help!


There's no general rule concerning that, as far as I know. Just follow the formula, and you won't go wrong! :P

Your line seems alright to me, since you've identified the fastest rate of change and where it occurs. Maybe you should find some examples. Perhaps ask a teacher if that's what is required from the question (whatever the question is).

You're welcome :-)
(edited 8 years ago)

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