Parametric Derivatives B01822

1. **State the problem:** We are given parametric equations $x = t^{2} + 2$ and $y = t^{2} + 9t$. We need to find the first derivative $\frac{dy}{dx}$, the second derivative $\frac{d^{2}y}{dx^{2}}$, and determine for which values of $t$ the curve is concave upward. 2. **Recall formulas:** - The first derivative for parametric equations is given by $$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ - The second derivative is $$\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$$ 3. **Calculate derivatives with respect to $t$:** - $\frac{dx}{dt} = \frac{d}{dt}(t^{2} + 2) = 2t$ - $\frac{dy}{dt} = \frac{d}{dt}(t^{2} + 9t) = 2t + 9$ 4. **Find $\frac{dy}{dx}$:** $$\frac{dy}{dx} = \frac{2t + 9}{2t}$$ 5. **Find $\frac{d}{dt}\left(\frac{dy}{dx}\right)$:** Let $u = 2t + 9$ and $v = 2t$, then $$\frac{dy}{dx} = \frac{u}{v}$$ Using the quotient rule: $$\frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^{2}}$$ Calculate derivatives: $$\frac{du}{dt} = 2, \quad \frac{dv}{dt} = 2$$ Substitute: $$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{2t \cdot 2 - (2t + 9) \cdot 2}{(2t)^{2}} = \frac{4t - 4t - 18}{4t^{2}} = \frac{-18}{4t^{2}}$$ 6. **Find $\frac{d^{2}y}{dx^{2}}$:** $$\frac{d^{2}y}{dx^{2}} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{-18/(4t^{2})}{2t} = \frac{-18}{4t^{2} \cdot 2t} = \frac{-18}{8t^{3}} = \frac{-9}{4t^{3}}$$ 7. **Determine concavity:** The curve is concave upward where $\frac{d^{2}y}{dx^{2}} > 0$. $$\frac{-9}{4t^{3}} > 0 \implies -9 > 0 \text{ if } t^{3} > 0$$ Since $-9$ is negative, for the fraction to be positive, $t^{3}$ must be negative. Therefore, the curve is concave upward when $$t^{3} < 0 \implies t < 0$$ **Answer:** $$\frac{dy}{dx} = \frac{2t + 9}{2t}$$ $$\frac{d^{2}y}{dx^{2}} = \frac{-9}{4t^{3}}$$ Concave upward interval: $$(-\infty, 0)$$