A few applications of calculus questions.. (1 Viewer)

[Makro]

Q16: The rate at which a silo fills with wheat is R = 6t tonnes/h. Initially the silo contains 3.2t.
a) How much wheat is in the silo after 7 hours?
b) When will the silo contain 50t?

The wording of this question is throwing me off. We're using T tonnes and t hours. So I'm not sure how to go about it. Answer for a) 150.2t b) 3.9h.

Q18: The temerpature T^oC of a compound in a science experiment is increasing exponentially over t minutes such that dT/dt = 0.03T. The temperature is initially 23^oC.
a) By what percentage is the temperature increasing each minute?
b) What is the temperature after 10 minutes?
c) When does the temperature reach 100^oC?

Again, here the wording is throwing me off. T temperature and t minutes. Just need the question explained differently I think. Answers: a) 3% b) 31^oC c) 49 mins

Q20: A particle is moving with acceleration a = 10e^2t ms^-2. If it's intially 4m to the right of the origin and has a velocity of -2 ms^-1, find it's displacement after 5 seconds to the nearest metre.

I attempted this a few times. I was just a few numbers off and had no idea what to do. Answer: 55 033 m.

All guidance would be much appreciated

 

[xFusion]

For question 20, the answer i got is 55033.66; i think its close enough, they probably rounded off early? anyways
a=10e^2t
v=5e^2t + C
when t=0, v=-2 ---> C= -7
v=5e^2t -7
x=(5/2)e^2t -7t +C
when t=0, x =4 ---> C=2.5
x=95/2)e^2t -7t +2.5
when t=5sec,
x=55033.66m

 

[simcon111]

Q16
a) You need to first integrate to get the displacement, meaning how much is in the silo at what time so integrate 6t.

Rate = 6t
Displacement = 3t^2 + C

You then work out the constant and you know when the time is 0 there is 3.2 tonns so C is 3.2

Displcement = 3t^2 + 3.2

So then you have to work out how much is in the silo at the 7 hour mark so you sub in 7 for t

Displacement at 7 hours: 3(7)^2 + 3.2
= 150.2

b) To work out when the silo will contain 50 tonnes you let the displcement = 50 and then solve for t

3t^2 + 3.2 = 50
3t^2 = 46.8
t^2 = 15.6
t = +- 3.9 (1dp)
but time can't be negative so t = 3.9

 

[xFusion]

Q18,
dT/dt=0.03T [flip]
dt/dT=1/(0.03T) [integrate]
t=(1/0.03)ln(0.03T) + C
initial temp =23; when t=0, T=23 ---> C=-(1/0.03)ln(0.03*23) =-(1/0.03)ln(0.69)
t=(1/0.03)[ln(0.03T) - ln(0.69)]
t=(1/0.03)ln(0.03T/0.69)
0.03t=ln(0.03T/0.69)
(0.69/0.03)e^(0.03t) = T
when t=10, T=31 degrees
when T=100 degrees, t=49min
for part a)
sub T=(0.69/0.03)e^(0.03t) into dT/dt=0.03T
---> dT/dt= 0.69e^(0.03t)
when t=0; dT/dt=0.69 ----1
when t=1; dT/dT =0.711 ----2
to find percentage put 2/1
0.711/0.69 =1.03
therefore 3% increase

 

[Timothy.Siu]

Q20: A particle is moving with acceleration a = 10e^2t ms^-2. If it's intially 4m to the right of the origin and has a velocity of -2 ms-1, find it's displacement after 5 seconds to the nearest metre.

a=dv/dt=10e^2t
v=5e^2t + C
t=0, v=-2
-2=5+C
C=-7

v=dx/dt=5e^2t-7
x=5/2 e^2t-7t +C
t=0 x=4
4=5/2+C
C=3/2
x=5/2 e^2t-7t+3/2
sub in t=5
x=5/2e^10-35+3/2
x=55033 (nearest metre)

 

[Makro]

Thanks everyone