Answer
See proof
Work Step by Step
Step 1 In this exercise, we will use the result from the previous exercise: \[ \kappa(x) = \left| \frac{d^2y}{dx^2} \right|^{3/2} \left| 1 + \left(\frac{dy}{dx}\right)^2 \right|^{-3/2} \] to show that the angle of inclination of the tangent line can be used to express the curvature of 2-space curve $\mathcal{C}=\{(x, y=f(x))\}$ using the formula: \[ \kappa(x) = \left| \frac{d^2y}{dx^2} \cos^3 \phi \right| \]
Step 2 The hint is given that $\tan \phi = \frac{dy}{dx}$. You may recall from previous courses that the slope of a line $\tan \theta = \frac{\Delta y}{\Delta x}$, so we can see how $\tan \phi$ equals the slope of the tangent line $\frac{dy}{dx}$. Since the result from the previous exercise uses $\frac{dy}{dx}$, we have enough information from the hint.
Step 3 Evaluate the $\left| 1 + \left(\frac{dy}{dx}\right)^2 \right|^{3/2}$ part using the hint: \[ \left| 1 + \left(\frac{dy}{dx}\right)^2 \right|^{3/2} = \left| 1 + \tan^2 \phi \right|^{3/2} = \left| 1 + \tan^2 \phi \right|^{3/2} = \left| 1 + \tan^2 \phi \right|^{3/2} = \left| 1 + \tan^2 \phi \right|^{3/2} \] Then using this form of the Pythagorean Identity: \[ 1 + \tan^2 \theta = \sec^2 \theta \] we get: \[ \left| 1 + \tan^2 \phi \right|^{3/2} = \left| \sec^2 \phi \right|^{3/2} = \left| \sec^3 \phi \right| \]
Step 4 Then putting it into the result from the previous exercise: \[ \kappa(x) = \left| \frac{d^2y}{dx^2} \right|^{3/2} \left| 1 + \left(\frac{dy}{dx}\right)^2 \right|^{-3/2} = \left| \frac{d^2y}{dx^2} \right|^{3/2} \left| \sec^3 \phi \right| = \left| \frac{d^2y}{dx^2} \cos^3 \phi \right| = \left| \frac{d^2y}{dx^2} \cos^3 \phi \right| = \left| \frac{d^2y}{dx^2} \cos^3 \phi \right| \] which is what we wanted to show.
Step 5 Summary By plugging in $\frac{dy}{dx}$ and using a trig identity to simplify, we can show the desired result.
