# Find \frac{dy}{dx} and \frac{d^2y}{dx^2} . \frac{dy}{dx}= \frac{d^2y}{dx^2}= For which...

## Question:

Find {eq}\frac{dy}{dx} {/eq} and {eq}\frac{d^2y}{dx^2} {/eq}.

{eq}\frac{dy}{dx}=\; \rule{20mm}{1pt} {/eq}

{eq}\frac{d^2y}{dx^2}=\; \rule{20mm}{1pt} {/eq}

For which values of {eq}t {/eq} is the curve concave upward? (Enter the answer using interval notation.)

{eq}\rule{20mm}{1pt} {/eq}

## Concavity of a Function:

{eq}\\ {/eq}

For a parametric function, we have : {eq}\dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} {/eq}, where {eq}x,y {/eq} are the functions of {eq}t {/eq}.

In order to find the interval in which the function is concave upwards, we need to find the value(s) of the variable for which its double derivative is greater than zero whereas for finding the interval in which the function is concave downwards, we need to find the value(s) of the variable for which its double derivative is negative.

{eq}\\ {/eq}

Let {eq}x(t)=4t^3+31, \ y(t)=3t^2-7t+89 {/eq}

(a) {eq}\dfrac{dy}{dx}=\dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} {/eq}

{eq}\Rightarrow \dfrac{dy}{dx}=\dfrac{6t-7}{12t^2} {/eq}

(b) {eq}\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{d}{dt}\left(\dfrac{6t-7}{12t^2}\right) {/eq}

{eq}\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{\left(12t^2.\dfrac{d}{dt}(6t-7)-(6t-7)\dfrac{d}{dt}12t^2\right)}{144t^4}\\\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{\left(6.12t^2-24t(6t-7)\right)}{144t^4}\\\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{\left(72t^2-144t^2+168t\right)}{144t^4}\\\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{\left(-72t^2+168t\right)}{144t^4}\\\Rightarrow \dfrac{d^2y}{dx^2}=\dfrac{-72t^2}{144t^4}+\dfrac{168t}{144t^4} {/eq}

(c) For the curve to be concave upwards, {eq}\Rightarrow \dfrac{d^2y}{dx^2}>0 {/eq}

{eq}\Rightarrow \dfrac{-72t^2+168t}{144t^4}>0\\\Rightarrow -72t^2+168t>0\\\Rightarrow -3t^2+7t>0\\\Rightarrow -3t(t-\dfrac{7}{3})>0\\\Rightarrow 0<t<\dfrac{7}{3} {/eq}

So, the interval for the graph being concave upwards is {eq}t\epsilon \left(0,\dfrac{7}{3}\right) {/eq}