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 :'(
/*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.
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
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.
To find the time at which the water mass is greatest, we need to determine the maximum value of the function. The maximum value of a function occurs at its vertex, which is the turning point where the function changes from increasing to decreasing.
To find the vertex of the function, we need to take the derivative of the mass function with respect to time, set it equal to zero, and solve for t.
Let's start by finding the derivative of the mass function:
m'(t) = 5*0.8*t^(0.8-1) - 3 = 4t^(-0.2) - 3
Setting m'(t) equal to zero:
4t^(-0.2) - 3 = 0
Solving for t, we get:
4t^(-0.2) = 3
t^(-0.2) = 3/4
(1/t)^(0.2) = 3/4
Taking the reciprocal of both sides:
t^(0.2) = 4/3
t = (4/3)^(1/0.2)
t = (4/3)^5
t ≈ 3.226 seconds
So, at approximately t = 3.226 seconds, the water mass is greatest.
To find the greatest mass, we can substitute this time value back into the original mass function:
m(t) = 5t^0.8 - 3t + 20
m(3.226) = 5*(3.226)^0.8 - 3*(3.226) + 20
m(3.226) ≈ 38.981 kilograms
Therefore, the greatest mass is approximately 38.981 kilograms.
Now, let's find the rate of mass change at t = 3 seconds and t = 5 seconds in kilograms per minute.
To find the rate of mass change, we need to find the derivative of the mass function with respect to time (m'(t)) and then convert it to kilograms per minute.
First, let's find m'(t) by differentiating the mass function:
m'(t) = 4t^(-0.2) - 3
To convert it to kilograms per minute, we need to divide the derivative by 60, as there are 60 seconds in a minute:
m'(3) = (4*(3)^(-0.2) - 3)/60
m'(5) = (4*(5)^(-0.2) - 3)/60
Evaluating these expressions, we get the rate of mass change at t = 3 seconds and t = 5 seconds in kilograms per minute.
Please note that I rounded the results to four decimal places:
m'(3) ≈ -0.0093 kg/min
m'(5) ≈ -0.0065 kg/min
Therefore, the rate of mass change at t = 3 seconds is approximately -0.0093 kilograms per minute, and at t = 5 seconds is approximately -0.0065 kilograms per minute.
only Time not the MAX Time
get*
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