consider the curve parametrized by (x = t^{4}+1) and (y = t^{6}+5t^{5}).\na) find (\frac{dy}{dx}) in terms…

Explanation:

Step1: Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$

Given $x = t^{4}+1$, then $\frac{dx}{dt}=4t^{3}$. Given $y=t^{6}+5t^{5}$, then $\frac{dy}{dt}=6t^{5}+25t^{4}$.

Step2: Calculate $\frac{dy}{dx}$ using the chain - rule $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

$\frac{dy}{dx}=\frac{6t^{5}+25t^{4}}{4t^{3}}=\frac{6t^{5}}{4t^{3}}+\frac{25t^{4}}{4t^{3}}=\frac{3}{2}t^{2}+\frac{25}{4}t$.

Step3: Find $\frac{d}{dt}(\frac{dy}{dx})$

$\frac{d}{dt}(\frac{dy}{dx})=\frac{d}{dt}(\frac{3}{2}t^{2}+\frac{25}{4}t)=3t + \frac{25}{4}$.

Step4: Calculate $\frac{d^{2}y}{dx^{2}}$ using the formula $\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$

$\frac{d^{2}y}{dx^{2}}=\frac{3t+\frac{25}{4}}{4t^{3}}=\frac{12t + 25}{16t^{3}}$.

Step5: Determine when the curve is concave - up

The curve is concave up when $\frac{d^{2}y}{dx^{2}}>0$. Set $\frac{12t + 25}{16t^{3}}>0$. We consider the sign of the numerator and denominator. The numerator $12t + 25 = 0$ when $t=-\frac{25}{12}$, and the denominator $16t^{3}=0$ when $t = 0$. We can use a sign - chart: If $t<-\frac{25}{12}$, let $t=-3$. Then $\frac{12(-3)+25}{16(-3)^{3}}=\frac{-36 + 25}{16\times(-27)}=\frac{-11}{-432}>0$. If $-\frac{25}{12}<t<0$, let $t =-\frac{1}{2}$. Then $\frac{12(-\frac{1}{2})+25}{16(-\frac{1}{2})^{3}}=\frac{-6 + 25}{16\times(-\frac{1}{8})}=\frac{19}{-2}<0$. If $t>0$, let $t = 1$. Then $\frac{12\times1+25}{16\times1^{3}}=\frac{37}{16}>0$.

Answer:

a) $\frac{dy}{dx}=\frac{3}{2}t^{2}+\frac{25}{4}t$ b) $\frac{d^{2}y}{dx^{2}}=\frac{12t + 25}{16t^{3}}$ c) $(-\infty,-\frac{25}{12}),(0,\infty)$