Respuesta :
A gray kangaroo jumped and it's path through the air is given by
y =[tex]-0.0267 x^2 +0.8 x[/tex]
where x is the kangaroo's horizontal distance traveled (in feet) wand y is the corresponding height (in feet).
For maximum distance we have to differentiate this expression.
[tex]\frac{\mathrm{d}y }{\mathrm{d} x}=-0.0267\frac{\mathrm{d} x^2}{\mathrm{d} x}+0.8\frac{\mathrm{d} x}{\mathrm{d} x}\\ \frac{\mathrm{d} y}{\mathrm{d} x}=-0.0267\times2x + 0.8[/tex]
For maxima or minima
[tex]\frac{\mathrm{d} y}{\mathrm{d} x}[/tex]=0
-0.0267×2 x + 0.8=0
⇒-0.0534 x + 0.8=0
⇒0.8=0.0534 x
⇒x=0.8/0.0534
x =14.98 (approx)
Now differentiating the expression again
[tex]\frac{\mathrm{d^{2}{y}} }{\mathrm{d} x^{2}}[/tex]=-0.0534
Since double derivative is negative , so
x=14.98 will be the point of Maxima.
The Kangaroo can go the maximum distance of 14.98 feet (approx)
and the height that can kangaroo go through=-0.0267×14.98×14.98+0.8×14.98
= -5.9914 +11.984
= 5.9925 feet
= 6 feet (approx)
So, X=Horizontal distance covered=14.98×2=29.96 feet,[∵ at x=14.98 it attains maximum height, So total distance Travelled by kangaroo will be two times of the distance covered that it has gone through to achieve maximum height.]⇒ The path is parabolic. = 30 feet (approx)
Y= Vertical distance covered= 5.9925=6 feet(approx)
