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Q. Consider the curve represented parametrically by the equation
$x=t^3-4 t^2-3 t \text { and } $
$y=2 t^2+3 t-5 \text { where } t \in R .$
If $H$ denotes the number of point on the curve where the tangent is horizontal and $V$ the number of point where the tangent is vertical then

Application of Derivatives

Solution:

$\frac{ dx }{ dt }=3 t ^2-8 t -3 ; \frac{ dy }{ dt }=4 t +3$
$\therefore \frac{ dy }{ dx }=\frac{4 t +3}{3 t ^2-8 t -3}$
Hence $H$ means $4 t +3=0 \Rightarrow t =-3 / 4 \Rightarrow H =1$
$V$ means $3 t^2-8 t-3=0 \Rightarrow t=3 \& t=-1 / 3 \Rightarrow V=2$