Question

Water is poured into a container that has a small leak.The
mass 'm' of the water is given as a function of time 't'
by,m=5t^0.8 -3t + 20 , with t>=0, t in seconds:

A) at what time is the water mass greatest,
B) what is the greatest mass?

*In kilogram per minute what is the rate of mass change at t=3 s, and t=5 s

plz answer me :'(

Answer 1

A) Time is ZERO T=0 because mass was in the max point before it started to leak.

B) substitute the max time to the function:

M= 5 x (0^0.8) - (3) x (0) + 20

Max MASS = 20g

c*) to gite the RATE of the mass we take derivative of the Function

M = (5) x (o.8) x (t^1-0.8) - 3 + 0

M = 4t^0.8 - 3

we just plug the time given to the new derived function :

Mass rate in T=3

(4) x (3^0.2) - 3 = 1.9

Mass rate in T=5

(4) x (5^0.2) - 3 = 2.5

Answer 2

only Time not the MAX Time

get*

Answer 3

/*At t=4.2 second mass will be MAXIMUM*/
Because at maxima slope needs to be zero and for this equation slope is zero (i.e horizontal to x axis when we plot graph of m(on y axis) vs t(on x axis)) at t=4.2seconds. You can do dm/dx and put it equal to zero and solve it you'll get t^0.2=3/4 or t=(3/4)^1/0.2 {write 1/0.2 in powers as 10/2 then 10/2=5, so t=(3/4)^5} then on solving we get value +-1024/243 (time can't be negative so take only positive value in consideration that is 1024/243 and that is equal to approximately 4.2.

Answer 4

Second part is easy just put t=4.2 in question.
Last part can be found by putting respective time values as asked in question into equation of dm/dt

Answer 5

Plot graph of 5x^0.8 - 3x + 20 = y (similar equation) on *DESMOS*(graph plot app/site) and zoom into1st quadrant(cuz we only have to check corresponding mass values for only positive values of time) and you will find Maxima at 4.2 second.