The curvature at a point P of a curve is defined as K = |dϕds| where ϕis the angle of inclinatio

step 1 In this question, we have a curvature question. We're told that the curvature at a point P of an

The curvature at a point P of a curve is defined as K =...

The curvature at a point $ P $ of a curve is defined as $$ K = \bigg|\dfrac{d\phi}{ds}\bigg| $$ where $ \phi $ is the angle of inclination of the tangent line at $ P $, as shown in the figure. Thus the curvature is the absolute value of the rate of change of $ \phi $ with respect to arc length. It can be regarded as a measure of the rate of change of direction of the curve at $ P $ and will be studied in greater detail in Chapter 13. (a) For a parametric curve $ x = x(t) $, $ y = y(t) $, derive the formula $$ K = \dfrac{|\dot{x}\ddot{y} - \ddot{x}\dot{y}|}{[\dot{x}^2 + \dot{y}^2]^{3/2}} $$ where the dots indicate derivatives with respect to $ t $, so $ \dot{x} = dx/dt $. [Hint: Use $ \phi = tan^{-1}(dy/dx) $ and Formula 2 to find $ d\phi/dt $. Then use the Chain Rule to find $ d\phi/ds $.] (b) By regarding a curve $ y = f(x) $ as the parametric curve $ x = x $, $ y = f(x) $, with parameter $ x $, show that the formula in part (a) becomes $$ K = \dfrac{|d^2y/dx^2|}{[1 + (dy/dx)^2]^{3/2}} $$

The curvature at a point $ P $ of a curve is defined as
$$ K = \bigg|\dfrac{d\phi}{ds}\bigg| $$
where $ \phi $ is the angle of inclination of the tangent line at $ P $, as shown in the figure. Thus the curvature is the absolute value of the rate of change of $ \phi $ with respect to arc length. It can be regarded as a measure of the rate of change of direction of the curve at $ P $ and will be studied in greater detail in Chapter 13.
(a) For a parametric curve $ x = x(t) $, $ y = y(t) $, derive the formula
$$ K = \dfrac{|\dot{x}\ddot{y} - \ddot{x}\dot{y}|}{[\dot{x}^2 + \dot{y}^2]^{3/2}} $$
where the dots indicate derivatives with respect to $ t $, so $ \dot{x} = dx/dt $. [Hint: Use $ \phi = tan^{-1}(dy/dx) $ and Formula 2 to find $ d\phi/dt $. Then use the Chain Rule to find $ d\phi/ds $.]
(b) By regarding a curve $ y = f(x) $ as the parametric curve $ x = x $, $ y = f(x) $, with parameter $ x $, show that the formula in part (a) becomes
$$ K = \dfrac{|d^2y/dx^2|}{[1 + (dy/dx)^2]^{3/2}} $$

Transcript

00:01 In this question, we have a curvature question.

00:03 We're told that the curvature at a point p of any curve is defined as the absolute value of d -fi ds.

00:12 And we're told that phi is the angle of inclination of the tangent line at a point p.

00:18 So it's told that it's the absolute value of the rate of change of phi for the respect to arc length.

00:24 So for example, if i have some arbitrary curve here, so let's suppose that our curve looks like that, and we have a point p, we have this tangent line here.

00:44 This angle here is phi.

00:48 So the curvature is rate of change of that angle.

00:54 We're asked to show a couple of different forms of curvature, first for a parametric curve, and then for curve as a function of x.

01:06 How can we do this? we're given a little hint.

01:11 We're told that phi is the arc tangent, of dydx, which obviously can be written as the arc tangent of d .ydt, dt, dt, dx, which can obviously again be rewritten to be the arc tangent of y dot over x dot.

01:45 We're using the hint earlier that x dot equals dxdt and y dot equals dydt.

01:51 And this is just obtained by division.

01:55 Now, obviously, the next thing we should do is we should do is we should.

01:59 Should take the derivative.

02:02 So then defy d t equals the time derivative of the arc tangent of y dot over x dot, which means obviously we're going to have to apply some chain rule.

02:21 Remembering that the derivative of the arc tangent is 1 over 1 plus x squared, we're going to have 1 plus y dot over x dot squared times the derivative of the inside.

02:39 Now the derivative of the inside, which we'll do down here, is if we square the denominator, we take the bottom, take the derivative of the top and subtract the top minus the derivative of the bottom.

03:03 Hence, this is what we have.

03:06 So we're going to have x, actually, we'll have x.

03:10 Y double dot minus y dot x double dot minus y dot x double dot and this whole denominator will be multiplied by x double dot squared now we can multiply this across and we're going to have that we're going to have that d phi d t would be equal to x.

03:34 Y double dot minus y dot x double dot over we have x .2 squared plus y dot squared now remember that remember the differential of arc length remember that it's the square root of x .d squared plus y dot squared d t, which means that obviously dsd would be just this quantity under the square root, which means that our equation for the curvature is equal to the absolute value of d5ds would be equal to, would be equal to d5dt, dtds.

04:34 And obviously that means we would have to reciprocate dtds.

04:35 And obviously that means we would have to reciprocate dtds...

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