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

YEEEEEET Mar 24, 2019

#3**0 **

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

CPhill Mar 25, 2019