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# minimum and maximum height

0
379
7 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

Mar 24, 2019

#1
0

You could try to find it by not using calculus, but it might be bashier.

Mar 24, 2019
#2
+1

If I'm reading this right, you just need to put t as 0 and 8.

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   Mar 25, 2019
#4
+1

LagTho  Mar 25, 2019
#5
-1

How did you find those numbers?

i subbed in 2 and 7 into the second differentiated equation but it wasnt those numbers, sorry im just really bad at these

YEEEEEET  Mar 25, 2019
#6
0

You need to put 2 and 7 into the original function to find the heights....sorry that I didn't make that clear...   Mar 25, 2019
#7
-1

thank you very much

YEEEEEET  Mar 25, 2019