Prove that the line is tangent to the curve at the point.

Question

Hello can someone please walk me through part a and b of the below question? I really want to understand it but am having a hard time figuring out the solution. I know how to calculate curvature for a parametrized and unparametrized curve using the respective formulas. Would I just take the derivatives of alpha, the respective cross products for the formula and simplify down to find the curvature? Thank you very much.

Answer 1

It should be understood that the polar coordinates are for the tangent point T or $ \alpha$ and not for the foot of perpendicular P.

I give a direct geometrical construction method based on differential lengths rather than a derivation.It is an attempt that is only partially successful,as of now, others may suggest improvements.

Separate sketch labels $(r, t) $ polar coordinate tangency point T and also $\phi$ and $\psi$.

From triangle OTX when external angle equated to sum of internal angles we get $$ t = \phi - \psi ...(1*) $$

By consideration of differential triangles, we can mark differentials $ dL, dp, p d\theta$ and $ q dL. $

Also $ d\theta = d \phi$ due to normal/tangent pair perpendicularity.

The radius r has components

$$ OP = p = r. sin \psi = dL/d\theta ... (2*) $$ and,

$$ PT = q = -r. cos\psi = dp / d\theta ...(3*) $$

along the normal and tangent respectively.

$$ x_T = p. cos\theta - q. sin \theta , y_T = p. sin\theta + q. cos \theta ... (4*) $$ Proposition 7 a).

Derivative of (1*) with respect to arc results in curvature

$$ kg = d\psi/dL + sin \psi/r ... (5*) $$

Plugging in from (2*) and (3*)

$$ kg= dp/dL/q ... (6*) $$

which does not agree with required result.

Arc Length $ L = \int p. d\theta $ Proposition 7 c).

but Area = $\int (p^2 + q^2)/2. d\theta $ which also does not agree with desired result.

EDIT2: The given result in proposition 7 d), if it does connote area in another way, most probably is a printing error of the sign.

EDIT1:

The example of a circle radius a above x-axis may be considered as counter-example for Proposition 7 b).

$p = OV + VC + CT = H/ sin \theta + h cos\theta + a = r\, sin \psi $

$ p' (\theta) = - H\, csc \theta\, cot \theta - h\, sin\theta $

$ p'' (\theta) = - H\, csc\theta\, ( csc^2\theta + cot^2\theta) - h\, cos\theta $

$ p (\theta) + p'' (\theta) = 2 H\, csc ^3\theta + a $

which does not agree with 7 b) unless H=0.