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Gradient of the graph y = a - k/x

  • Thread starter DeanBH
  • Start date
  • #1
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" find, in terms of a and k, the gradient of the graph y = a - k/x at the point where it crosses x axis."


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
 

Answers and Replies

  • #2
nicksauce
Science Advisor
Homework Helper
1,272
5
I imagine you would want to calculate dy/dx at that point by plugging in x = k/a into your expression for dy/dx.
 
  • #3
737
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How would you normally calculate the value of the derivative dy/dx at a point x?
 
  • #4
cristo
Staff Emeritus
Science Advisor
8,107
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Why are you all calculating derivatives? This is in the precalculus forum :confused:
 
  • #5
737
0
The OP used derivatives, and it looks like a calculus problem. Maybe it should be moved?
 
  • #6
82
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didn't think it was worthy of the higher forum

i don't know how to simplify it properly, that's the problem. =(
 
  • #7
82
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can someone tell me how to simplify the answer, this isn't homework or coursework!

just revising need help! thanks
 
  • #8
ah i think it should be:

[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
82
0
ah i think it should be:

[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
thought so, cheers beef
 

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