The Student Room Group

Equations of tangents and normals

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Where am I going wrong :cry2:
Reply 1
Original post by IFoundWonderland


Where am I going wrong :cry2:


The line after m=m = \cdots how did you get the four in the denominator?

It should be m=3÷242×22m = 3 \div \frac{2}{4} - \frac{2 \times 2}{2}
Original post by Zacken
The line after m=m = \cdots how did you get the four in the denominator?

It should be m=3÷242×22m = 3 \div \frac{2}{4} - \frac{2 \times 2}{2}

:undefined:facepalm

Thanks :lol:
Reply 3
Original post by IFoundWonderland
Thanks :lol:


No problem.
Original post by Zacken
No problem.


Hi, sorry to bother you.

I don't really understand this :s-smilie:

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Reply 5
Original post by IFoundWonderland
Hi, sorry to bother you.


If I write dxx2\int \frac{\mathrm{d}x}{x^2} I really mean 1x2dx\int \frac{1}{x^2} \, \mathrm{d}x - the former notation just saves a bit of space.

Anyhow, you know how to intgrate 4h2dh=4h2dh\int \frac{4}{h^2} \, \mathrm{d}h = 4\int h^{-2} \, \mathrm{d}h using the adding one the power and dividing by the new power, etc...
Original post by Zacken
If I write dxx2\int \frac{\mathrm{d}x}{x^2} I really mean 1x2dx\int \frac{1}{x^2} \, \mathrm{d}x - the former notation just saves a bit of space.

Anyhow, you know how to intgrate 4h2dh=4h2dh\int \frac{4}{h^2} \, \mathrm{d}h = 4\int h^{-2} \, \mathrm{d}h using the adding one the power and dividing by the new power, etc...


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Oh yay, thanks so I did it right :woo:
Reply 7
Original post by IFoundWonderland
...


Well done. :smile:

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