A curve is defined by the parametric equations $x=t^{2}-16$ and $y=t^{4}+3 t$.

What is $\frac{d^{2} y}{d x^{2}}$ in terms of $t ?$
Choose 1 answer:
(A) $2-\frac{3}{4 t^{3}}$
(B) $2 t^{2}+\frac{3}{2 t}$
(C) $t+\frac{3}{4 t^{2}}$
(D) $\frac{2 t}{4 t^{3}+3}$

Question

A curve is defined by the parametric equations $x=t^{2}-16$ and $y=t^{4}+3 t$.

What is $\frac{d^{2} y}{d x^{2}}$ in terms of $t ?$
Choose 1 answer:
(A) $2-\frac{3}{4 t^{3}}$
(B) $2 t^{2}+\frac{3}{2 t}$
(C) $t+\frac{3}{4 t^{2}}$
(D) $\frac{2 t}{4 t^{3}+3}$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find $\frac{d^{2} y}{d x^{2}}$ in terms of $t$, we first need to find $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$. From the asksia-ll calculator result, we have $\frac{dy}{dt} = 4t^3 + 3$ and $\frac{dx}{dt} = 2t$. ‖ ‖ step 2 ⋮ Now, $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t^3 + 3}{2t}$. ‖ ‖ step 3 ⋮ Next, we differentiate $\frac{dy}{dx}$ with respect to $t$ to find $\frac{d}{dt}\left(\frac{dy}{dx}\right)$. This gives us $\frac{d}{dt}\left(\frac{4t^3 + 3}{2t}\right)$. ‖ ‖ step 4 ⋮ Applying the quotient rule, $\frac{d}{dt}\left(\frac{4t^3 + 3}{2t}\right) = \frac{(12t^2)(2t) - (4t^3 + 3)(2)}{(2t)^2}$. ‖ ‖ step 5 ⋮ Simplifying, we get $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{24t^3 - 8t^3 - 6}{4t^2} = \frac{16t^3 - 6}{4t^2}$. ‖ ‖ step 6 ⋮ Finally, $\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} = \frac{\frac{16t^3 - 6}{4t^2}}{2t} = \frac{16t^3 - 6}{8t^3} = 2 - \frac{3}{4t^3}$. ‖ ***A*** ∻Key Concept∻ ⚹Second derivative of a parametric curve⚹ ∻Explanation∻ ⚹To find the second derivative of a parametric curve, we differentiate the first derivative of $y$ with respect to $x$ ($\frac{dy}{dx}$) with respect to $t$, and then divide by $\frac{dx}{dt}$.⚹◊What is the derivative of $y$ with respect to $x$ when $x=t^{2}-16$ and $y=t^{4}+3t$?⍭ Generate me a similar question◊

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