for this question since i think minimum and maximum height of a curve is calculated by differentiate the curve and make it equal to 0
i did:
\(dt/ds= \frac{4t^3 -36t^2 +56t}{50}, \frac{4t^3 -36t^2 +56t}{50} =0, t= 0\)
which didnt work as t should be max= 8.64 min= 1.14
please help thanks
ds/dt = [ 4t^3 - 36t^2 + 56t ] / 50 = 0
(4/50) ( t^3 - 9t^2 + 14t) = 0
(4/50)t [ t^2 - 9t + 14) = 0
t ( t - 7) (t - 2) = 0
t = 0 t = 2 or t =7
Take the second derivative of [t^3 - 9t^2 + 14t ]
[3t^2 - 18t + 14 ]
At t = 0...this is positive...so....a rel min and the height= 8 ft
At t =2 ....this is 0
At t =1, the first derivative is positive
At t =3, the first derivative is negative
So...a t= 2...we have a max.....and the height is 8.64 ft
At t = 7 ....this is 0
At t = 8....the first derivative is positive
So...at t = 7...we have another min and the height is 1.14 ft
So.....
max = (2, 8.64)
min = ( 7, 1,14 )
Here's the graph to confirm this : https://www.desmos.com/calculator/rdjmso6wrw