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Yeolsi
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#1
Hey! Any help would be very much appreciated 
a) Investigate the gradient of the curve y=x^3+3 at the points X=1 and X=2. Suggest a formula for the gradient function of this curve.
b) Use the technique of differentiation from first principles to verify the formula you suggested.
a) Investigate the gradient of the curve y=x^3+3 at the points X=1 and X=2. Suggest a formula for the gradient function of this curve.
b) Use the technique of differentiation from first principles to verify the formula you suggested.
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Notnek
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#2
(Original post by kimmylee)
Hey! Any help would be very much appreciated
a) Investigate the gradient of the curve y=x^3+3 at the points X=1 and X=2. Suggest a formula for the gradient function of this curve.
b) Use the technique of differentiation from first principles to verify the formula you suggested.
Hey! Any help would be very much appreciated
a) Investigate the gradient of the curve y=x^3+3 at the points X=1 and X=2. Suggest a formula for the gradient function of this curve.
b) Use the technique of differentiation from first principles to verify the formula you suggested.
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Yeolsi
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#3
(Original post by Notnek)
What have you tried so far? How you're meant to do the "investigation" in a) depends on the context. Is this from a textbook?
What have you tried so far? How you're meant to do the "investigation" in a) depends on the context. Is this from a textbook?
It says use a spreadsheet to investigate it but there's a way to calculate the values using a table without doing it on a spreadsheet but I'm not sure how it works
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Notnek
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#4
(Original post by kimmylee)
Hey!
It says use a spreadsheet to investigate it but there's a way to calculate the values using a table without doing it on a spreadsheet but I'm not sure how it works
Hey!
It says use a spreadsheet to investigate it but there's a way to calculate the values using a table without doing it on a spreadsheet but I'm not sure how it works
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Yeolsi
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#5
(Original post by Notnek)
Is this a textbook question? Can you post a picture of it?
Is this a textbook question? Can you post a picture of it?
It's not from a textbook, it's my homework I was given and that's kind of all it says
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Notnek
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#6
(Original post by kimmylee)
Hey! It's not from a textbook, it's my homework I was given and that's kind of all it says![Name: image-3386d265-2c1d-424d-acc0-a28dcfe291021542755456-compressed.jpg.jpeg
Views: 80
Size: 63.0 KB]()
Hey!
It's not from a textbook, it's my homework I was given and that's kind of all it saysI'm guessing that for a) you need to investigate the gradients of chords (lines joining two points on the curve) close to x = 1. The smaller the chord, the better the approximation will be for the actual gradient at x = 1.
So e.g. you could look at x = 1 and x = 1.2 (a point close to 1) and work out the y values for each and find the gradient of this chord.
So at x = 1, y = 1^3 + 3 = 4
At x = 1.2, y = 1.2^3 + 3 = 4.728
Then the gradient is (y2 - y1) / (x2 - x1) = (4.728 - 4)/(1.2 - 1) = 3.64.
Then try with x = 1 and x = 1.1. Then try even closer etc. and this will get closer and closer to the true gradient at x = 1. You can use Excel techniques to speed this up.
If you've been taught to do it differently then let us know.
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Yeolsi
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#7
(Original post by Notnek)
You should know about the context of this better than anyone here because you were in the lesson
I'm guessing that for a) you need to investigate the gradients of chords (lines joining two points on the curve) close to x = 1. The smaller the chord, the better the approximation will be for the actual gradient at x = 1.
So e.g. you could look at x = 1 and x = 1.2 (a point close to 1) and work out the y values for each and find the gradient of this chord.
So at x = 1, y = 1^3 + 3 = 4
At x = 1.2, y = 1.2^3 + 3 = 4.728
Then the gradient is (y2 - y1) / (x2 - x1) = (4.728 - 4)/(1.2 - 1) = 3.64.
Then try with x = 1 and x = 1.1. Then try even closer etc. and this will get closer and closer to the true gradient at x = 1. You can use Excel techniques to speed this up.
If you've been taught to do it differently then let us know.
You should know about the context of this better than anyone here because you were in the lesson
I'm guessing that for a) you need to investigate the gradients of chords (lines joining two points on the curve) close to x = 1. The smaller the chord, the better the approximation will be for the actual gradient at x = 1.
So e.g. you could look at x = 1 and x = 1.2 (a point close to 1) and work out the y values for each and find the gradient of this chord.
So at x = 1, y = 1^3 + 3 = 4
At x = 1.2, y = 1.2^3 + 3 = 4.728
Then the gradient is (y2 - y1) / (x2 - x1) = (4.728 - 4)/(1.2 - 1) = 3.64.
Then try with x = 1 and x = 1.1. Then try even closer etc. and this will get closer and closer to the true gradient at x = 1. You can use Excel techniques to speed this up.
If you've been taught to do it differently then let us know.
Omg, thank you so much. And yh, that's how i did it in lesson but they asked it in a different way and I got really confused cos it didn't mention specific coordinates and finding the gradient. Anyway, thanks a lot, I think I'll be able to do it now
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Notnek
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#8
(Original post by kimmylee)
Hey Omg, thank you so much. And yh, that's how i did it in lesson but they asked it in a different way and I got really confused cos it didn't mention specific coordinates and finding the gradient. Anyway, thanks a lot, I think I'll be able to do it now
Hey
Omg, thank you so much. And yh, that's how i did it in lesson but they asked it in a different way and I got really confused cos it didn't mention specific coordinates and finding the gradient. Anyway, thanks a lot, I think I'll be able to do it now
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