Consider the curve defined by the given parametric equations.

Find the first derivative (in terms of ).

Find the second derivative (in terms of ).

Determine whether the curve is concave up or concave down at
Concave down
Concave up

Ask by Hilton Bernard. in the United States

Dec 14,2024

Question

Consider the curve defined by the given parametric equations.

Find the first derivative (in terms of ).

Find the second derivative (in terms of ).

Determine whether the curve is concave up or concave down at
Concave down
Concave up

Ask by Hilton Bernard. in the United States

Dec 14,2024

UpStudy AI · Accepted Answer

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 $$\frac{\pi}{2} < t < \frac{3\pi}{2}$$. 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 curve is concave down when $$0 < t < \frac{\pi}{2}$$ or $$\frac{3\pi}{2} < t < 2\pi$$, and concave up when $$\frac{\pi}{2} < t < \frac{3\pi}{2}$$.

Consider the curve defined by the given parametric equations.

Find the first derivative (in terms of ).

Find the second derivative (in terms of ).

Determine whether the curve is concave up or concave down at
Concave down
Concave up

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Consider the curve defined by the given parametric equations. Find the first derivative (in terms of ). Find the second derivative (in terms of ). Determine whether the curve is concave up or concave down at Concave down Concave up

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Consider the curve defined by the given parametric equations.

Find the first derivative (in terms of ).

Find the second derivative (in terms of ).

Determine whether the curve is concave up or concave down at
Concave down
Concave up