Consider the curve defined by the given parametric equations. \( \left\{\begin{array}{l}x=-5+\sec (t) \\ y=-4+2 \tan (t)\end{array}\right. \) Find the first derivative (in terms of \( t \) ). \( \frac{d y}{d x}= \) Find the second derivative (in terms of \( t \) ). \( \frac{d^{2} y}{d x^{2}}= \) Determine whether the curve is concave up or concave down at \( t= \) Concave down Concave up

To find the first derivative \(\frac{dy}{dx}\) in terms of \(t\), we start by finding \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\). Given the parametric equations: \[ x = -5 + \sec(t) \] \[ y = -4 + 2\tan(t) \] ### Step 1: Find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) 1. **Differentiate \(x\) with respect to \(t\)**: \[ \frac{dx}{dt} = \frac{d}{dt}(-5 + \sec(t)) = \sec(t) \tan(t) \] 2. **Differentiate \(y\) with respect to \(t\)**: \[ \frac{dy}{dt} = \frac{d}{dt}(-4 + 2\tan(t)) = 2\sec^2(t) \] ### Step 2: Find \(\frac{dy}{dx}\) Using the chain rule, we have: \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2\sec^2(t)}{\sec(t) \tan(t)} \] Simplifying this expression: \[ \frac{dy}{dx} = \frac{2\sec^2(t)}{\sec(t) \tan(t)} = \frac{2\sec(t)}{\tan(t)} = 2 \cdot \frac{1}{\cos(t)} \cdot \frac{\cos(t)}{\sin(t)} = \frac{2}{\sin(t)} \] Thus, the first derivative is: \[ \frac{dy}{dx} = \frac{2}{\sin(t)} \] ### Step 3: Find the second derivative \(\frac{d^2y}{dx^2}\) To find the second derivative, we need to differentiate \(\frac{dy}{dx}\) with respect to \(t\) and then divide by \(\frac{dx}{dt}\). 1. **Differentiate \(\frac{dy}{dx}\) with respect to \(t\)**: \[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{2}{\sin(t)}\right) = -\frac{2\cos(t)}{\sin^2(t)} \] 2. **Now, divide by \(\frac{dx}{dt}\)**: \[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{-\frac{2\cos(t)}{\sin^2(t)}}{\sec(t) \tan(t)} \] Substituting \(\sec(t) = \frac{1}{\cos(t)}\) and \(\tan(t) = \frac{\sin(t)}{\cos(t)}\): \[ \frac{d^2y}{dx^2} = \frac{-\frac{2\cos(t)}{\sin^2(t)}}{\frac{1}{\cos(t)} \cdot \frac{\sin(t)}{\cos(t)}} = \frac{-2\cos^2(t)}{\sin^2(t) \cdot \frac{\sin(t)}{\cos(t)}} = \frac{-2\cos^3(t)}{\sin^3(t)} \] Thus, the second derivative is: \[ \frac{d^2y}{dx^2} = \frac{-2\cos^3(t)}{\sin^3(t)} \] ### Step 4: Determine concavity at a specific \(t\) To determine whether the curve is concave up or concave down, we look at the sign of \(\frac{d^2y}{dx^2}\). - If \(\frac{d^2y}{dx^2} > 0\), the curve is concave up. - If \(\frac{d^2y}{dx^2} < 0\), the curve is concave down. Since \(\frac{d^2y}{dx^2} = \frac{-2\cos^3(t)}{\sin^3(t)}\), the sign depends on \(\cos(t)\) and \(\sin(t)\): - If \(\cos(t) > 0\) and \(\sin(t) > 0\) (i.e., \(0 < t < \frac{\pi}{2}\)), then \(\frac{d^2y}{dx^2} < 0\) (concave down). - If \(\cos(t) < 0\) and \(\sin(t) > 0\) (i.e., \(\frac{\pi}{2} < t < \pi\)), then \(\frac{d^2y}{dx^2} > 0\) (concave up). - If \(\sin(t) < 0\), then \(\frac{d^2y}{dx^2}\) will also be negative regardless of \(\cos(t)\). ### Conclusion - The first derivative is: \[ \frac{dy}{dx} = \frac{2}{\sin(t)} \] - The second derivative is: \[ \frac{d^2y}{dx^2} = \frac{-2\cos^3(t)}{\sin^3(t)} \] - The concavity depends on the value of \(t\): - Concave down if \(0 < t < \frac{\pi}{2}\) or \(\frac{3\pi}{2} < t < 2\pi\). - Concave up if \(\frac{\pi}{2} < t < \frac{3\pi}{2}\).