- #1

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ok i worked out dy/dx = k/x^2 and x = k/a when y = o.

now what do i do. =(

thx for help in advance

- Thread starter DeanBH
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- #1

- 82

- 0

ok i worked out dy/dx = k/x^2 and x = k/a when y = o.

now what do i do. =(

thx for help in advance

- #2

nicksauce

Science Advisor

Homework Helper

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- #3

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How would you normally calculate the value of the derivative dy/dx at a point x?

- #4

cristo

Staff Emeritus

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Why are you all calculating derivatives? This is in the precalculus forum

- #5

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The OP used derivatives, and it looks like a calculus problem. Maybe it should be moved?

- #6

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i don't know how to simplify it properly, that's the problem. =(

- #7

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just revising need help! thanks

- #8

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[tex]\frac{dy}{dx}[/tex] = -[tex]\frac{k}{x^{2}}[/tex] = -[tex]\frac{k}{(\frac{k}{a})^{2}}[/tex] = - [tex]\frac{k}{\frac{k^{2}}{a^{2}}}[/tex] = -[tex]\frac{ka^{2}}{k^{2}}[/tex] = -[tex]\frac{a^{2}}{k}[/tex]

when y = 0

- #9

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thought so, cheers beef

[tex]\frac{dy}{dx}[/tex] = -[tex]\frac{k}{x^{2}}[/tex] = -[tex]\frac{k}{(\frac{k}{a})^{2}}[/tex] = - [tex]\frac{k}{\frac{k^{2}}{a^{2}}}[/tex] = -[tex]\frac{ka^{2}}{k^{2}}[/tex] = -[tex]\frac{a^{2}}{k}[/tex]

when y = 0

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