question 7\n10 pts\ngiven: ( x = 1 + t ^ { 2 }, y = t - t ^ { 3 })\na) find the values of ( t ) that…

question 7\n10 pts\ngiven: ( x = 1 + t ^ { 2 }, y = t - t ^ { 3 })\na) find the values of ( t ) that corresponds to horizontal and vertical tangent lines to the graph of ( c ). (you just need to find ( t ). you do not need to find ( ( x, y ) ).)\nb) find ( \frac { d ^ { 2 } y } { d x ^ { 2 } } ) in terms of ( t ).\nshow all steps clearly. enter your final answer.\n(a) horizontal tangent at: \nveritical tangent at:\n(b) \nedit view insert format tools table\n12pt paragraph b i u a t ^ { 2 }

question 7\n10 pts\ngiven: ( x = 1 + t ^ { 2 }, y = t - t ^ { 3 })\na) find the values of ( t ) that corresponds to horizontal and vertical tangent lines to the graph of ( c ). (you just need to find ( t ). you do not need to find ( ( x, y ) ).)\nb) find ( \frac { d ^ { 2 } y } { d x ^ { 2 } } ) in terms of ( t ).\nshow all steps clearly. enter your final answer.\n(a) horizontal tangent at: \nveritical tangent at:\n(b) \nedit view insert format tools table\n12pt paragraph b i u a t ^ { 2 }

Answer

Explanation:

Step1: Find (\frac{dx}{dt}) and (\frac{dy}{dt})

For (x = 1+t^{2}), (\frac{dx}{dt}=2t). For (y=t - t^{3}), (\frac{dy}{dt}=1 - 3t^{2}).

Step2: Find (t) for horizontal tangent

A horizontal tangent occurs when (\frac{dy}{dt}=0) and (\frac{dx}{dt}\neq0). Set (\frac{dy}{dt}=0), so (1 - 3t^{2}=0). [ \begin{align*} 3t^{2}&=1\ t^{2}&=\frac{1}{3}\ t&=\pm\frac{1}{\sqrt{3}} \end{align*} ] When (t = \pm\frac{1}{\sqrt{3}}), (\frac{dx}{dt}=2t\neq0).

Step3: Find (t) for vertical tangent

A vertical tangent occurs when (\frac{dx}{dt}=0) and (\frac{dy}{dt}\neq0). Set (\frac{dx}{dt}=0), so (2t = 0), then (t = 0). When (t = 0), (\frac{dy}{dt}=1-3(0)^{2}=1\neq0).

Step4: Find (\frac{d^{2}y}{dx^{2}})

We know that (\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{1 - 3t^{2}}{2t}). Using the quotient rule (\frac{d}{dt}(\frac{u}{v})=\frac{u^{\prime}v - uv^{\prime}}{v^{2}}) where (u = 1-3t^{2}), (u^{\prime}=-6t) and (v = 2t), (v^{\prime}=2). (\frac{d}{dt}(\frac{1 - 3t^{2}}{2t})=\frac{-6t\times(2t)-(1 - 3t^{2})\times2}{(2t)^{2}}=\frac{-12t^{2}-2 + 6t^{2}}{4t^{2}}=\frac{-6t^{2}-2}{4t^{2}}). And (\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}) [ \begin{align*} \frac{d^{2}y}{dx^{2}}&=\frac{\frac{-6t^{2}-2}{4t^{2}}}{2t}\ &=\frac{-6t^{2}-2}{8t^{3}}\ &=\frac{-3t^{2}-1}{4t^{3}} \end{align*} ]

Answer:

(a) Horizontal tangent at: (t=\pm\frac{1}{\sqrt{3}}) Vertical tangent at: (t = 0) (b) (\frac{d^{2}y}{dx^{2}}=\frac{-3t^{2}-1}{4t^{3}})