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/sci/ - Science & Math


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File: 1.48 MB, 2423x2423, Eagle Nebula M16.jpg [View same] [iqdb] [saucenao] [google]
2202824 No.2202824 [Reply] [Original]

Starting out with differential calculus here... Got this bombshell dropped on me and I have no clue how to solve it, so any nudges in the right direction would be wholly appreciated.

Points P1 (1, y1), P2, (1.01, y2) and P3 (1.1, y3) lie on the curve y = kx^2. THe gradient of the chord P1 P3 is 6.3 and the gradient of the chord P1P2 is 6.03.

What do these results suggest about the gradient of the tangent to the curve y = kx^2 at P1?

Also, I'm asked to deduce K.

Go on /sci/, show off your maths-fu.

pic unrelated

>> No.2202845

just took a final exam over all this shit. im praying to god to let me get a C.

>> No.2202855

>>2202845

any clue on how to approach this then?

I differentiated the kx^2 to dy/dx = 2kx

So at (1, y1), dy/dx = 2k(1) or 2k

Doesn't seem very useful?

>> No.2202873
File: 447 KB, 1280x2590, the fairy of Eagle Nebula.jpg [View same] [iqdb] [saucenao] [google]
2202873

bump

>> No.2202897

LOOKS LIKE DY/DX MIGHT BE 6x OP GEE I WONDER WHAT K COULD BE IF IT MULTIPLIES BY 2 TO MAKE 6

>> No.2202931

>>2202897

How do you arrive at 6x?

An explanation would be better than spending the same of time on unhelpful trolling, to be honest. I'm asking for your help.

>> No.2202943

>>2202931
As the difference between the chords gets smaller(infinitely), the gradient appears to approach 6 (at x=1). Seeing as x=1 we can come to the conclusion that dy/dx = 6x.
To find 'K' just integrate. If you don't know how to do that, just do the reverse of differentiation. If you don't know the general formula for differentiation, you're not far into differentiation so you should learn it soon.
Is this college level stuff in America?

>> No.2202949

maybe cosmoleaks.org knows more about the nebula...but whats the point with the points??

>> No.2202969

>>2202943

I saw this link but I wasn't sure if that was a valid way of working this out.

So if dy/dx = 6x

and y = kx^2 differentiated is
dy/dx = 2kx
then since they are equal 2kx should = 6x?

so 2kx = 6x

K = 3

I checked the mark scheme and that's right. Thanks a lot!