anonymous
  • anonymous
at what points on the curve does y = x^3 slope at 45 degrees?
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
The gradient function is the derivative of your equation. You'd have,\[y=x^3 \rightarrow y'=3x^2\]Now, the slope of a line is equal to the tangent that line makes with the positive horizontal axis. A slope of 45 degrees is then,\[m=\tan 45 ^o=1\]Since y' gives the gradient at any x, you have then,\[y'=1 \rightarrow 3x^2=1\]You then solve for x.
anonymous
  • anonymous
\[x=\pm \frac{1}{\sqrt{3}}\]
anonymous
  • anonymous
oh i see now, i didn't realize that a slope of 45 degrees meant tan 45. thankx a million lokisan!

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anonymous
  • anonymous
np :)

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