$\begingroup$Find the value of $\frac{d^2y}{dx^2}$ at the point by the given value of t. $$ x=9t^2-6, y=t^5, t=1 $$ I am way off when I calculate this. As I understand it $\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
and $\frac{d^2y}{dx^2}=\frac{\frac{dy'}{dt}}{\frac{dx}{dt}}$
I get y' = $\frac{20t^3}{18t}$ and then $y'' = \frac{20^3}{18}$ but this gives me an answer of $\frac{4000}{9}$ which is obviously wrong.
What am I doing wrong here?
$\endgroup$03 Answers
$\begingroup$First, let's differentiate $x$ and $y$ with respect to $t$:
- $x = 9t^2 -6 \Rightarrow \frac{dx}{dt} = 18t$ and $\frac{dt}{dx} = \frac{1}{18t}$
- $y = t^5 \Rightarrow \frac{dy}{dt} = 5t^4$
Now $\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dt} \left( \frac{dy}{dx} \right)\frac{dt}{dx} = \frac{d}{dt} \left(\frac{dy}{dt} \frac{dt}{dx} \right) \frac{dt}{dx}$
Inserting the values, we obtain
$\frac{d}{dt} \left(\frac{5t^4}{18t} \right) \frac{1}{18t}= \frac{5}{324 t} \frac{d}{dt}\left(t^3 \right) = \frac{5 \cdot 3}{324}t$
For $t= 1$ the result is therefore $\frac{d^2 y }{dx^2 } = \frac{5}{108}$
$\endgroup$1$\begingroup$$$\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)\div\frac{dx}{dt}=\left[\frac{\frac{d^2y}{dx^2}\frac{dx}{dt}-\frac{dy}{dt}\frac{d^2x}{dt^2}}{\left(\frac{dx}{dt}\right)^2}\right]\div\frac{dx}{dt}=\frac{(20t^3)(18t)-(5t^4)(18)}{(18t)^3}=\frac{5t}{108}$$
$\endgroup$$\begingroup$Firstly x'=18t and y'=5t^4 Hence answer will be 5t/108 Hence answer at t=1 is 5/108
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