# one more question

• Sep 18th 2007, 09:11 PM
bungy189
one more question
I have this questions an answer to it would be great.
.Q.1 At t minutes after an engine starts, the rate of fuel consumption is Rkg per minute. it is given by R=10+ 10/2t+1.
a) Sketch R as a function of t
b) find the rate of consumption after 7 mins
c) find the limiting value of R as t becomes very large
d) calculate the total consumption in the 1st 7 mins
not sure how to do it
Thank You
• Sep 19th 2007, 03:11 AM
ticbol
Quote:

Originally Posted by bungy189
I have this questions an answer to it would be great.
.Q.1 At t minutes after an engine starts, the rate of fuel consumption is Rkg per minute. it is given by R=10+ 10/2t+1.
a) Sketch R as a function of t
b) find the rate of consumption after 7 mins
c) find the limiting value of R as t becomes very large
d) calculate the total consumption in the 1st 7 mins
not sure how to do it
Thank You

R = 10 +[10 /(2t +1)]

a) Sketch R as a function of t
So, vertical axis is for R; horizontal axis is for t.
Ordered pair is (t,R).

When t = 0, R = 10 +[10/1] = 20, so, (0,20)
When t = 1, R = 10 +[10/3] = 13.333, so, (1,13.333)
When t = 2, R = 10 +[10/5] = 12, so, (3,12)
Umm, it is not linear, because the differences foy R are not the same.
When t = 10, R = 10 +[10/21] = 10.476, so, (10,10.476)

I don't know how to draw here, but the the graph is a smooth curve that starts at (0,20) and decreasing as t becomes larger.
You can do the sketch on paper. Plot all those solved ordered pairs or points, then run a smooth curve through them.
The limit of the curve as t grows bigger is the horizontal line R = 10.

b) find the rate of consumption after 7 mins
R7 = 10 +[10 /(2*7 +1)] = 10 +[10/15] = 10.667 kg/min.

c) find the limiting value of R as t becomes very large
As mentioned in part (a), R = 10 kg/min.

d) calculate the total consumption in the 1st 7 mins
Umm, integration.

Consumption, C = R*t
C7 = INT.(0-->7)[R*dt]
C7 = INT.(0-->7)[10 +10/(2t +1)] dt
C7 = (10)INT.(0-->7)[1 +1/(2t +1)]dt
C7 = 10[t +(1/2)ln(2t +1)]|(0-->7)
C7 = 10{[7 +(1/2)ln(15)] -[0 +(1/2)ln(0+1)]
C7 = 10[7 +(1/2)ln(15)]