Question

A function y=f(x) is represented parainetrically as follows: $x=\varphi \left(t\right)=t^5-5t^3-20t+7.$
$y=\Psi \left(t\right)=4t^3-3t^2-18t+3.$ Let the values of "t" at extrema of this function be t1 and t2.Find 2(t1+t2)? Note: $(-2<t<2)$

Answer 1

$x=\Phi \left(t\right)=t^5-5t^3-20t+7$
$\therefore \dot{x}=5t^4-15t^2-20$$=5(t^4-3t^2-4)=5(t^2-4)(t^2+1)$
Clearly $\dot{x}\ne 0$ when $-2<t<2$ ...(1)
$y=\Psi \left(t\right)=4t^3-3t^2-18t+3$
$\dot{y}=12t^2-6t-18=6(2t^2-t-3)$$=6(2t-3)(t+1)$
$\therefore \dot{y}=0$ gives $t=-1,3/2$
Both these values satisfy the condition
$-2<t<2$ ...(2)
Now $\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}=0(x\ne 0)$
$\therefore \dot{y}=0\therefore t=-1,3/2$ by (1) and (2)
$\therefore \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{\dot{y}}{\dot{x}}\right)=\frac{d}{dt}\left(\frac{\dot{y}}{\dot{x}}\right)\frac{dt}{dx}$
$=\frac{\dot{x}\ddot{y}-\dot{y}\ddot{x}}{\dot{x^2}}.\frac{1}{\dot{x}}=\frac{\dot{x}\ddot{y}}{\dot{x^2}}.\frac{1}{\dot{x}}$
$\because y=0$ and $x\ne 0$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{\ddot{y}}{\dot{x^2}}=\frac{6(4t-1)}{\dot{x^2}}<0$ at $t=-1$ and $>0$ at $t=3/2$
Hence $y$ is max. at $t=-1($ i.e. $y=14,x=31)$ and is minimum at $t=3/2$
$(i.e.,y=-17\frac{1}{4},x=-\frac{1033}{32}).$