- The displacement s of a piston during each 8-s is given by s = 8t -t^2. For what value of t is the velocity of the piston 4?
- The distance s travelled by a subway train after the brakes are applied is given by s = 20t -2t^2 How far does it travel after the brakes are applied in coming to a stop?
- Water is being drained from a pond such that the volume V of water in the pond after t hours is given by V = 50t(60-t^2). Find the rate at which the pond is being drained after 4 hours.
- The electric field E at a distance r from a point charge is E=k/r^2 where k is a constant. Find an expression for the instantaneous rate of change of the electric field with respect to r.
- The altitude h of a certain rocket as a function of the time t after launching is given by h = 550t - 4.9 t^2. What is the maximum altitude the rocket attains?
- The blade of a saber saw moves vertically up and down and its displacement is given by y = 1.85 sin t. Find the velocity of the blade for t=0.2
- Differentiate y = 7sin x + 4x^2 +1
Sunday, December 6, 2009
Real Life ie rate of change
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s=8t-t^2
ReplyDeleteds/dt= 8-2t
when s = 4
4= 8-2t
4-8= -2t
-4= -2t
-4/-2= t
2=t
please tell me if im wrong i kinda try ah ting here
Your final answer is correct BUT the working is not quite correct
ReplyDeletes=8t-t^2
WHAT IS ds/dt rate of change of distance wrt time
hat's right, velocity so differentiating is the way to go
ds/dt= 8-2t
Here is where you went wrong
when s = 4 when dist = 4 BUT dist is not given but let say it was why then did you not put dist as 4 you said s=4 but what you actually did was put ds/dt = 4 so from here you would be given wrong which is unfortunate since you final answer is correct
when s = 4
4= 8-2t
4-8= -2t
-4= -2t
-4/-2= t
2=t
i doh understand miss.. what exactly was i supposed to do then?
ReplyDeleteZoe stay calm you can do it
ReplyDeletePractice the differentiation of polynomial
you did do that correct
you error was question said velocity = 4
and v = ds/dt
But you said the wrrong thing but did the correct thing
you will do well
s=8t-t^2
ReplyDeletethen ds/dt= 8-2t
so when s=4
suds into the equ.
4=8-2t
2t=8-4
2t=4
therefore t=4/2
=2
so then t is 2
that was number one
ReplyDeletes = 20t -2t^2
ReplyDeletes=20-4t
i doh no what to do now! miss HELP!!!!
#1
ReplyDeletedisplacement = s
s=8t-t^2
i.e using loop
ds/dt=8-2t
i.e when ds/dt=4 sub.
4=8-2t
8-2t=4
-2t=8-4
t=8-4/2
t=4/2
i.e t=2
#2
ReplyDeletes=20t-2t^2
using loop
ds/dt=20-4t
#1
ReplyDeletedisplacement = s
s=8t-t^2
ds/dt=8-2t
i.e when ds/dt=4
4=8-2t
8-2t=4
-2t=8-4
t=8-4/2
t=4/2
t=2
This comment has been removed by the author.
ReplyDelete2. s = 20t -2t^2
ReplyDeleteds/dt = 20 - 4t
adn for the second part, don't we need to know time(t)?
1. s = 8t -t^2
ReplyDeleteds/dt = 8 - 2t
when ds/dt = 4
4 = 8 - 2t
8 - 2t = 4
-2t = -4
t = -4/-2
t = 2
s=20t-2t^2
ReplyDeleteds/dt=20-4t when ds/dt=0
0= 20-4t
t= -20/4
t= -5
sub t= -5
s= 20(-5)-2(-5)^2
s= -100 - 50
s= -150
distance is cannot be negative
therefore s= 150
p.s. my answer could be wrong.... help me if you can!
4. E=k/r^2
ReplyDeleteE = (k)r^-2
dE/dr = -2k/r^3
snowwhite?? what exactly is a loop?
ReplyDeletefor no.3
ReplyDeletev=50t(60-t^2)
multiply out your brackets
v= 3000t-50t^3
differentiate v wrt t
dv/dt= 3000-150t^2
sub t=4 into dv/dt
dv/dt= 3000- 150(4)^2
dv/dt= 3000- 150(16)
dv/dt= 3000-2400= 600
question 1
ReplyDeletedisplacement = s
s=8t-t^2
ds/dt=8-2t
i.e when ds/dt=4
4=8-2t
8-2t=4
-2t=8-4
t=8-4/2
t=4/2
t=2
yea serious......what is the loop?
ReplyDeletefor no. 4
ReplyDeleteE=k/r^2
=kr^-2
dE/dr= -2kr^-3
5. h = 550t - 4.9t^2
ReplyDeletedh/dt = 550 - 9.8t
following the method of "d blackman"...
when dh/dt = 0
550 - 9.8t = 0
-9.8t = -550
t = 56.12
susbt t = 56.12
h = 550t - 4.9t^2
h = 550(56.12) - 4.9*(56.12^2)
h = 30866 - 15432.32656
h = 15433.67344
h = 15433.7
∴ maximum height = 15433.7
This comment has been removed by the author.
ReplyDelete7newstarlight7.. please explain how you got the max altitude
ReplyDeletey = 1.85 sin t
ReplyDeletedy/dt = 1.85 cos t
when t = 0.2
dy/dt = 1.85 x cos (0.2)
dy/dt = 1.85
Well zoe as i said in my comment, i did what d blackman did. He first solved for 't' which happens to be the same variable in this equation as in his, but yeah solve for 't' and substitute the value you got into the original equation. Just like when you want to find the 'y' value at an 'x' point in any normal equation. First you find the 'x' value or values, in this case one value because 550 - 9.8t = 0 is not a quadratic equation and substitute them into your original equation. It's just that in this case its 'h' which is normally 'y' and 't' which is normally 'x'. Does that help any?
ReplyDeleteP.S. sorry for using the word equation so much. And also i have to say that neither me nor d blackman are 1000000000000% sure that this is correct but from a logical standpoint it certainly seems so.
7. y = 7sinx + 4x^2 +1
ReplyDeletedy/dx = 7cosx + 8x
This comment has been removed by the author.
ReplyDeletenumber1:
ReplyDeletes=8t-t^2
then ds/dt= 8-2t
so when s=4
sub s=4 into the eqn.
4=8-2t
2t=8-4
2t=4
t=4/2
=2
therefore t is 2
@snow white for number 2
ReplyDeletei understand the first 2 lines... hahhahaa
but WHAT IS LOOP? why you randomly used it to get your answer?
This comment has been removed by the author.
ReplyDeletesorry about the above post... i was totally wrong.
ReplyDeletenumber3:
v=50t(60-t^2)
expand brackets
v= 3000t-50t^3
dv/dt= 3000-150t^2
sub t=4 into dv/dt
dv/dt= 3000- 150(4)^2
dv/dt= 3000- 150(16)
dv/dt= 3000-2400
= 600
#7
ReplyDeletey = 7sin x + 4x^2 +1
dy/dx of sin x = cos x;
dy/dx = 7 cos x + (2)4x^(2-1)
dy/dx = 7 cos x + 8x
#4
ReplyDeleteE = k/r^2
E = k(r)^-2
dE/dr = (-2)k(r)^(-2-1)
dE/dr = -2k(r)^-3
dE/dr = -2k/r^3
#3
ReplyDeleteV = 50t(60-t^2).
V = 3000t - 50t^3
although this is out of the syllabus as the highest power that can be asked is 2(am i right?cant remember...i believe so though);
dV/dt = 3000 - (3)50t^(3-1)
dV/dt = 3000 - 150t^2
when t=4;
dV/dt = 3000 - 150(4)^2
dV/dt = 3000 - 150(16)
= 3000 - 2400
dV/dt = 600
#1
ReplyDeletedisplacement = s
1)
s=8t-t^2
ds/dt=8-2t
v=ds/dt then, v=4
4=8-2t
8-2t=4
-2t=8-4
t=4/2
t=2
2)
ReplyDeletes=20t-2t^2
ds/dt= 20-4t
ds/dt=v
v=20-4t
at a stop then v=0
0=20-4t
4t=-20
t=-5
assuming - just gives direction
s=20(-5)-2(-5)^2
s= -150
dist travelled b4 stop = 150
no5.
ReplyDeleteh = 550t - 4.9 t^2
dh/dt = 550 - 2(4.9)t
for max pts dh/dt =0
=> 550 -9.8t = 0
9.8t=550
t= 550/9.8
t = 56.1
at t=56.1 the rocket will have max altitude
sub 56.1 into h
= 550 (56.1) -4.9(56.1)^2
max altitude = 15433.7 units
no.6 y = 1.85 sin t
ReplyDeletedy/dt= 1.85 cos t
where t= 0.2
1.85cos(0.2)
v =1.85
#3
ReplyDeleteV = 50t(60-t^2).
V = 3000t - 50t^3
dV/dt = 3000 - (3)50t^(3-1)
dV/dt = 3000 - 150t^2
when t=4;
dV/dt = 3000 - 150(4)^2
dV/dt = 3000 - 150(16)
= 3000 - 2400
dV/dt = 600
(4)4. E=k/r^2
ReplyDeleteE = (k)r^-2
dE/dr = -2k/r^3
1). s=8t-t^2
ReplyDeleteds/dt=8-2t
given that ds/dt=4
8-2t=4
-2t=4-8
-2t=-4
t=-4/-2
t=2
2). s=20t-2t^2
ReplyDeleteds/dt=20-4t
to find t ds/dt=0
20-4t=0
-4t=-20
t=-20/-4
t=5
when t=5
s=20(5)-2(5)^2
s=100-50
s=50
3).V=50t(60-t^2)
ReplyDeleteV=3000t-50t^3
dV/dt=3000-150t^2
when t=4hours
dv/dt=3000-150(4)2
dv/dt=3000-2400
dv/dt=600
4). E=k/r^2
ReplyDeleteE=kr^-2
dE/dr= -2kr^-3
5). h=550t-4.9t^2
ReplyDeletedh/dt=550-9.8t
for max. height dh/dt=0
550-9.8t=0
-9.8t=-550
t=-550/-9.8
t=56.12
when t= 56.12
h=550(56.12)-4.9(56.12)^2
h=30866-15432.33
h= 15433.67
6).
ReplyDeletey= 1.85 sin t
dy/dt= 1.85 cos t
when t=0.2
dy/dt= 1.85 cos 0.2
dy/dt= 1.85
7). y=7 sinx+4x^2+1
ReplyDeletedy/dx= 7cosx+8x
no.1
ReplyDeletedisplacement = s =8t-t^2
ds/dt=8-2t
(when ds/dt=4)
4=8-2t
8-2t=4
-2t=8-4
t=8-4/2
t=4/2
t=2
ques 7
ReplyDeletey = 7 sinx + 4x^2 + 1
dy/dx = 7cosx + 2(4x)
dy/dx = 7cosx + 8x
question 3
ReplyDeleteV = 50t(60 - t^2)
V = 3000t - 50t^3
dv/dt = 3000 - 150t^2
when t=4
dv/dt= 3000 - 150(4)^2
dv/dt = 3000 - 150(16)
dv/dt = 3000 - 2400
dv/dt = 600
the volume of water in the pnd will be changing at a rate of 600 after 4 hours
yup that max altitude thing is 100% correct, just checked it out
ReplyDelete1)
ReplyDeletes=8t-t^2
ds/dt=8-2t
v=ds/dt then, v=4
4=8-2t
8-2t=4
-2t=8-4
t=4/2
t=2
question 1
ReplyDeletes=8t-t^2
ds/dt=8-2t
when ds/dt=4
4=8-2t
4-8=-2t
-4=-2t
-4/-2=t
t=2 sec
question 2
ReplyDeletes=20t-2t^2
ds/dt=20-4t
when ds/dt=0
20-4t=0
20=4t
20/4=t
t=5
substitute t=5 in s=20t-2t^2
s=20(5)-2(5)^2
s=100-2(25)
s=100-50
s=50
question 3
ReplyDeleteV=50t(60-t^2)
V=3000t-50t^3
dV/dt=3000-150t^2
when t=4
dV/dt=3000-150(4)^2
=3000-2400
=600
This comment has been removed by the author.
ReplyDeletequestion 4
ReplyDeleteE=k/r^2
E=kr^-2
dE/dr=-2kr^-3
=-2k/r^3
question 5
ReplyDeleteh=500t-4.9t^2
dh/dt=500-9.8t
dh/dt=0
500-9.8t=0
500=9.8t
500/9.8=t
51.02=t
substitute t=51.02 in h=500t-4.9t^2
h=500(51.02)-4.9(51.02)^2
=25510-12754.90
=12755.1
question 6
ReplyDeletey=1.85sint
dy/dx=1.85cost
when t=0.2
dy/dx=1.85cos(0.2)
=1.85
to bookworm referring to question 2
ReplyDelete(copying your wrking)
question 2
s=20t-2t^2
ds/dt=20-4t
when ds/dt=0
20-4t=0
20=4t
20/4=t
t=5
substitute t=5 in s=20t-2t^2
s=20(5)-2(5)^2
s=100-2(25)
s=100-50
s=50
when u stated ds/dt =0
and u had
20-4t=0
ur next step said
20/4=t
when it should have been
-20/4=t
question 7
ReplyDeletey=7sinx+4x^2+1
dy/dx=7cosx+8x
This comment has been removed by the author.
ReplyDeletenumber5:
ReplyDeleteh=550t-4.9t^2
dh/dt=550-9.8t
max. height/ altitude dh/dt=0
550-9.8t=0
-9.8t=-550
t=-550/-9.8
t=56.12
when t= 56.12
h=550(56.12)-4.9(56.12)^2
h=30866-15432.33
h= 15433.67
number6:
ReplyDeletey= 1.85 sin t
dy/dt= 1.85 cos t
when t=0.2
dy/dt= 1.85 cos 0.2
= 1.85
number7:
ReplyDeletey=7sinx+4x^2+1
dy/dx=7cos+[4x^2-1]
dy/dx=7cosx+8x
1.
ReplyDeletes=8t - t^2
ds/dt = 8 - 2t
when ds/dt=4
4=8-2t
-2t=-4
t=2
2.
ReplyDeletes = 20t -2t^2
ds/dt=20-4t
when ds/dt=0
0=20-4t
-20=-4t
t=5
subs. t=5 into s = 20t -2t^2
s = 20(5) - 2(5)^2
s= 100 - 50
s=50
3.
ReplyDeleteV = 50t(60-t^2).
v= 3000t - 50t^3
dv/dt= 3000 - 150t^2
when t=4
dv/dt = 3000 - 150(4)^2
dv/dt = 3000 - 2400
dv/dt = 600
4.
ReplyDeleteE=k/r^2
de/dr = -2kr^-3
5.
ReplyDeleteh = 550t - 4.9 t^2
dh/dt = 550 - 9.8t
when dh/dt = 0
0 = 550 - 9.8t
-9.8t = -550
t = 56.12
subs. t int0,h = 550t - 4.9 t^2
h = 550(56.12) - 4.9(56.12)^2
h = 30866 - 15432.32
h = 15433.68
6.
ReplyDeletey = 1.85 sin t
dy/dt = 1.85 cos t
when t=0.2
dy/dt = 1.85cos.2
dy/dt = 1.85
#3
ReplyDeletev=50t(60-t^2)
v=3000t-50t^2
i.e dv/dt=3000-150t^2
7.
ReplyDeletey = 7sin x + 4x^2 +1
dy/dx = 7cos x + 8x
#3
ReplyDeletewhen t=4 sub
dv/dt=3000-150(4)^2
dv/dt=3000-150(16)
dv/dt=3000-2400
dv/dt=600
#4
ReplyDeleteE=K/r^2
E=kr^-2
de/dr=-2kr^-3
#5
ReplyDeleteh=550t-4.9t^2
dv/dt=550-9.8t
#5
ReplyDeletewhen dv/dt=0
550-9.8t=0
-9.8t=-550
t=-550/-9.8
i.e t=56.12
#5
ReplyDeletesub when t=56.12
h=550t-4.9t^2
h=550(56.12)-4.9(56.12)^2
h=30866-15432.3
h=1543.7
#6
ReplyDeletey=1.85 sine t
dy/dt=1.85 cos
#6
ReplyDeletesub when t=0.2
dy/dt=1.82 cos(0.2)
dy/dt=1.85
#7
ReplyDeletey=7sin+4x^2+1
dy/dt=7cosx+8x
Q1.
ReplyDeletes = 8t -t^2
ds/dt = 8 - 2t
v = ds/dt
When v = 4 :
4 = ds/dt
therefore
4 = 8 - 2t
2t = 8 - 4
2t = 4
t = 2
Q2.
ReplyDeleteds/dt = speed
s = 20t -2t^2
ds/dt = 20 - 4t
When ds/dt = 0 the train will stop
hence:
0 = 20 - 4t
4t = 20
t = 5
The distance s travelled by a subway train after the brakes are applied is given by s = 20t -2t^2 How far does it travel after the brakes are applied in coming to a stop?
ReplyDeletes=20t-2t^2
ds/dt=20-4t
ds/dt=0
0=20-4t
4t=20
t=20/4
t=5
Water is being drained from a pond such that the volume V of water in the pond after t hours is given by V = 50t(60-t^2). Find the rate at which the pond is being drained after 4 hours.
v=50t(60-t^2)
v=(50t)(60)- (50t)(t^2)
v=3000t-50t^3
dv/dt=3000-150t^2
:.=3000-150(4)^2
dv/dt=3000-150(16)
dv/dt=3000-2400
dv/dt=600
The blade of a saber saw moves vertically up and down and its displacement is given by y = 1.85 sin t. Find the velocity of the blade for t=0.2
Differentiate y = 7sin x + 4x^2 +1
y=7sin x+4x^2+1
dy/dx= 7cos x +8x
no.3
ReplyDeletes=20t-2t^2
ds/dt=20-4t at stop ds/dt= 0
0= 20-4t
t= -20/4
t= 5
sub t= 5
s= 20(5)-2(5)^2
s= 100 - 50
s= 50
therefore s= 50
no.3
ReplyDeletes=20t-2t^2
ds/dt=20-4t at stop ds/dt= 0
0= 20-4t
t= -20/4
t= 5
sub t= 5
s= 20(5)-2(5)^2
s= 100 - 50
s= 50
therefore s= 50
= 50t(60 - t^2)
ReplyDeleteV = 3000t - 50t^3
dv/dt = 3000 - 150t^2
when t=4
dv/dt= 3000 - 150(4)^2
dv/dt = 3000 - 150(16)
dv/dt = 3000 - 2400
dv/dt = 600
the volume of water in the pnd will be changing at a rate of 600 after 4 hours
This is for question 3 is this correct ????
ds/dt = speed
ReplyDeletes = 20t -2t^2
ds/dt = 20 - 4t
When ds/dt = 0 the train will stop
hence:
0 = 20 - 4t
4t = 20
does this cover the rate of brakes appliesd
t = 5
2.
ReplyDeletes= 20t -2t^2
ds/dt=20 - 4t
4.
ReplyDeleteE=k/r^2
E=kr^-2
dE/dr= -2k/r^3
s=8t-t^2
ReplyDeletethen ds/dt= 8-2t
so when s=4
suds into the equ.
4=8-2t
2t=8-4
2t=4
therefore t=4/2
=2
= 50t(60 - t^2)
ReplyDeleteV = 3000t - 50t^3
dv/dt = 3000 - 150t^2
when t=4
dv/dt= 3000 - 150(4)^2
dv/dt = 3000 - 150(16)
dv/dt = 3000 - 2400
dv/dt = 600
the volume of water in the pnd will be changing at a rate of 600 after 4 hours
7. diff.
ReplyDeletey =7sinx+4x^2 +1
dy/dx=7cosx+8x
i think sum1 did this similarly but i worked it out and got reletivly the same thing
ReplyDelete