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#### finchie_88

##### Guest

I've just started learning about intrinsic equations after learning about envelopes, arc length and curved surface area in cartesian, parametric and polar coordinates, and understand the principle behind intrinsic equations, but my book doesn't explain it particularly well, here is a question I'm stuck on and my attempt so far:

For the curve [tex] y= \ln|secx| [/tex] prove that [tex]\frac{ds}{dx} = secx [/tex] and that [tex]\psi = x [/tex]. Hence find the intrinsic equation.

[tex]y = \ln|secx| \therefore \frac{dy}{dx} = tanx[/tex]

[tex] \frac{ds}{dx} = \frac{ds}{dy} x \frac{dy}{dx} [/tex]

[tex] \frac{ds}{dx} = \sqrt{1 + (\frac{dx}{dy})^2} = \frac{tanx.secx}{tanx} = secx [/tex]

I don't know how to prove that [tex] x = \psi [/tex], but I think for the last part that [tex] s = \int sec \psi . d \psi [/tex], is this correct?

Can someone explain or show me if what I have so far is right, and how to complete the question? any help would be appreciated.

edit: Where it says ds/ds = ds/dy x dy/dx, the x in the middle is suppose to be a multiply sign.

**Question:**For the curve [tex] y= \ln|secx| [/tex] prove that [tex]\frac{ds}{dx} = secx [/tex] and that [tex]\psi = x [/tex]. Hence find the intrinsic equation.

**My answer (Very incomplete):**[tex]y = \ln|secx| \therefore \frac{dy}{dx} = tanx[/tex]

[tex] \frac{ds}{dx} = \frac{ds}{dy} x \frac{dy}{dx} [/tex]

[tex] \frac{ds}{dx} = \sqrt{1 + (\frac{dx}{dy})^2} = \frac{tanx.secx}{tanx} = secx [/tex]

I don't know how to prove that [tex] x = \psi [/tex], but I think for the last part that [tex] s = \int sec \psi . d \psi [/tex], is this correct?

Can someone explain or show me if what I have so far is right, and how to complete the question? any help would be appreciated.

edit: Where it says ds/ds = ds/dy x dy/dx, the x in the middle is suppose to be a multiply sign.

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