- 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 = 50(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 voltage V induced in an inductor in an electric circuit is given by V = L(d^2q/dt^2) where L is the inductance. Find the expression for the voltge induced in a 1.6 H inductor if q = (2t + 1)^.5 -1.
- 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 36∏t. Find the velocity of the blade for t=0.025.
- The charge q on a capacitor in a circuit containing a capacitor of capacitance C, a resistance R, and a source of voltage Eis given by q = CE(1 - e^(-t/RC) ). Show that this equation satisfies the equation Rdq/dt + q/C = E.
- An earth orbiting satellite is launched such that its altitude is given by y = 240(1 - e^(-.05t)). Find the velocity of the satellite for t= 10.
- Differentiate y = 7sin x + ln (4x^2 +1)
Thursday, March 26, 2009
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given the displacement (s) =8t-t^2
ReplyDeletev = d/t
we are given s and v
v = 4
we can differentiate the equation s = 8t-t^2
ds/dt= 8 -2t
since ds/dt = velocity = 4
we can solve for the time
4 = 8 -2t
t = -4/-2
t = 2
for question #2
ReplyDeletes is given by s= 20t -2t^2
ds/dt = 20 -4t
since ds/dt= velocity and the velocity of the train after applying brakes = 0 then
0=20-4t
therefore t= -20/-4
t=5
the time the train takes to come to a stop after the brakes are applied is 5.
therefore the distance it travels can now be found by simply substituting t=5 into the equation which represents The displacement traveled by a subway train after the brakes are applied is s= 20t -2t^2
therefore
s=(20*5) -2(5^2)
s=100 - 50
s=50
s = 8t -t^2
ReplyDeleteds/dt = 8t - 2t
The value of the pistion is 4 when t is
4 = 8t - 2t
making t the subject
t = 8/ -4 / -2t
t = -4 / -2t
t = 2
This comment has been removed by the author.
ReplyDeleteE=k/r^2
ReplyDeleteBring all downstairs upstairs
E=kr^-2
DE/DR
E=2kr^-3
(s) =8t-t^2
ReplyDeletevelocity is ds/dt, the change of distance with respect to time.
ds/dt = 8 - 2t when ds/dt = 4
4 = 8 - 2t
4 - 8 = - 2t
-4 = - 2t
t = 2
y = 7sin x + ln (4x^2 +1)
ReplyDeletesub m = (4x^2 +1)
dm/dx = 8x
y = ln m
dy/dm = 1/m
dy/dx = dy/dm x dm/dx
= 1/m x 8x
= (1/4x^2 +1)x 8x
= 8x/4x^2 +1
therefore dy/dx = 7cos x + (8x/4x^2 +1)
q.2
ReplyDeletedisplacement s = 8t-t^2
v=d/t
given that v=4
s=8t-t^2
therefore...
ds/dt=8-2t
t=-4/-2
t=2
Q9 ;
ReplyDeletevelocity is the change in distance by the change in time
altitude is the distance
so velocity = dy/dt
but there is notted terms so we
remove the brackets to get y=240-240e^(-.05t)
but we have -.05t complicated
we let u= -.05t
then du/dt = -.05
we rewrite the original equation to get
y=240-240e^u
we know dy/dx e^x = e^x
so dy/du = -240e^u
then dy/dt = dy/du * du/dt
-240e^(-.05t)*-.05
there are some errors to question 4 by low rider
ReplyDeleteyes we bring downstairs to upstairs
but there is now a police in the new equation
E=Kr^-2
now if u use y=ax^n ; dy/dx = (a*n)x^n-1
so multiply k by -2 gives -2K
so the final answer should be
dE/dr not E
so dE/dr = -2Kr^-3
Q3.
ReplyDeleteIgnore the background. look at the equation given and differientiate because the ques is asking for the rate at which...
V = 50(60-t)^2
let m = 60-t dm/dt = -1
rewrite:
V = 50m^2 dV/dm = 100m
The ques asked for dV/dt:
dV/dt = dm/dt x dV/dm
dV/dt = (-1) x 100(60-t)
When t = 4
dV/dt = (-1) x 100(60-4) = -100 x 56 = -5600
Can someone confirm if im correct??
Question 1
ReplyDeletes = 8t-t^2
v = d/t
v = 4
differentiate the equation s = 8t-t^2
ds/dt= 8 -2t
since ds/dt = v = 4
then t can be found
4 = 8 -2t
t = -4/-2
t = 2
Question 2
ReplyDeletes = 20t -2t^2
ds/dt = 20 -4t
since ds/dt = v = 0
then t can be found
0 = 20-4t
t = -20/-4
t = 5
The time the train takes to come to a stop after the brakes are applied is 5.
Simply substitut t=5 into the equation which represents the displacement traveled which is s = 20t -2t^2 to find the distance travelled
Thus
s = (20)(5) - 2(5^2)
s = 100 - 50
s = 50
Yes "Dark Angel" i believe the answer you have for question 3 is correct because i did it also and got the same answer...
ReplyDeleteQuestion 4
ReplyDeleteAn expression for the instantaneous rate of change of the electric field with respect to r.
E = k/r^2
E = kr^-2
dE/dr = -2kr^-3
Question 6:
ReplyDeleteh = 550t + 4.9t^2
dh/dt = 550 + 9.8t
since h is the altitude of the rock, the maximum altitude the rock reaches will be:
0 = 550 + 9.8t
9.8t = 550
t = 550/9.8
=56.12
Ans = 56.12
This comment has been removed by the author.
ReplyDeleteQuestion 10
ReplyDeletey = 7 sin x + ln (4x^2 +1 )
It looks a lil confusing so,
let s = (4x^2 +1 )
therefore
ds/dx = 8x
y = ln s
dy/ds = 1/s
thus we multiple the both new terms to find dy/dx which will give:
dy/dx = dy/ds x ds/dx
= 1/s x 8x
= (1/4x^2 + 1)x 8x
= 8x/4x^2 + 1
therefore the final answer for dy/dx
= 7 cos x + (8x/4x^2 + 1)
Can someone help me out with questions 5, 7, 8, and 9 plez. I tried them all but keep getting stock.. But question 7 i dont even know how to start..
ReplyDeleteI really need the help....
for question 7
ReplyDeletewe first ignore the background and look for a equation
which we see is y=1.85sin∏t
this from the question is said to be the displacement
now they want the velocity
and we know velocity is displacement divided by time
so dy/dt
but we only know dy/dx of sin x = cosx
but we have ∏t complicated
so we let m=∏t
dm/dt = ∏
rewrite y= 1.85sin(m)
dy/dm = 1.85cos(m)
but we want dy/dt
so dy/dt = dm/dt * dy/dm
= (∏)*(1.85cos(m))
we replace ∏t for m we get
1.85∏cos∏t
we then replace 0.025 for t to get the final ans
question 5
ReplyDeletefor the first equation they give us
they say V = L(d^2q/dt^2)
if you look further
we get a second equation q = (2t+1)^.5 - 1
now if we find dg/dt
the d^2q/dt^2 in the first equation just mean
we diff a second time
i have done question 9 so swanky thing
ReplyDeleteyou can check and see if it helps
question 8 is difficult
ReplyDeletei may need help with this one
can some one do it if at least part of it
read some comments here an this is how i would work the first question:
ReplyDeletes= 8t-t^2
therefore
ds/dt= 8-2t
solving for the time:
4=8-2t
4-8=-2t
-4=-2t
therefore
2=t
read some more posts after attempting question 2 got it out after:
ReplyDeletethe velocity after the applied brakes is 0
then find for t, then sub the value for t into the original eq'n
2nd question goes like this:
ReplyDeletes=20t-2t^2
therefore
ds/dt=20-4t
velocity after brakes is applied =0
therefore
0=20-4t
20/4=t
therefore
5=t
now that t was found
sub t=5 into the original eq'n
s=(20*5)-2(5^2)
s=100-50
s=50
For question 7 what does ∏ mean in the question. Goldfinger probly you could tell me.
ReplyDeleteQ1.
ReplyDeletes = 8t-t^2
v = d/t
v = 4
differentiate the equation s = 8t-t^2
ds/dt= 8 -2t
since ds/dt = v = 4
FOR t=
4 = 8 -2t
t = -4/-2
t = 2
Water is being drained from a pond such that the volume V of water in the pond after t hours is given by V = 50(60-t)^2. Find the rate at which the pond is being drained after 4 hours.
ReplyDeleteFind the rate at which the pond is being drained is another way of asking for differentiation
therefore
dv/dt =
this is where i have problems please help me
Joh's..question
ReplyDeleteequation..from..problem
v = 50(60-t)^2
let m = 60-t dm/dt = -1
so
i..get
v = 50m^2
dv/dm = 100m
The question asked for dV/dt:
dv/dt = dm/dt x dV/dm
dv/dt = (-1) x 100(60-t)
When t = 4
dv/dt = (-1) x 100(60-4)
dv/dt= -100 x 56
applications`of`differentiation`&`integration
ReplyDeleteThe`velocity`v`ms^-1,of`a`particle`moving`in`a
straight`line`,t`seconds`after`passing`through`a`
fixed`point`O`
is`given`by:
v=6t^2-5t+3.
Find:
1)the`acceleration`of`the`particle`when`t=3.
2)the`distance`of`the`particle`from`O`when`t=4.
solution.....
ReplyDelete1)To`find`acceleration:dv/dt
note:dv/dt`is`the`rate`of`change`of`velocity`with
respect`to`time,so,therefore`dv/dt=acceleration
dv/dt=12t-5
when`t=3
dv/dt=12(3)-5
dv/dt=31
so`acceleration=31ms^-2
2)answer=100m
The`displacement`from`O`is`the`area`under`the
v-t`graph..
could`someone`help`me`with`question`10..
ReplyDeletequestion 10
ReplyDeletey= 7sin x +ln (4x^2+1)
first let m=4x^2+1 so that there are no brackets and no square terms
then take the each one of the terms m, sin and ln and differentiate
differentiate m
dm/dx= 8x
differentiate sin
y=7 sin x
dy/dx=-7 cos x
differentiate ln
y= ln m
dy/dm= 1/m
the question asked for dy/dx so to egt this dm/dx * dy/dm
=8x*1/m=8x/m
dy/dx=-7cos x+ 8x/m
place back what is m because you changed it to differentiate
Ans= -7 cosx+ 8x/4x^2+1
weezy i answerred question 10 to the best of my knowledge of differentiation
ReplyDeleteA rectangle box is made from a piece of cardboard 8cm by 12 cm by cutting equal square x cm, from each corner and bending up the sides. Express the volume of the box as a function of the side of the square that is cut out.
ReplyDeletethe question above have given me some problems some one please help me
ReplyDeleteAn earth orbiting satellite is launched such that its altitude is given by
ReplyDeletey = 240(1 - e^(-.05t)). Find the velocity of the satellite for t= 10.
finding the letter for (1- e^(-0.05t)to put?
can some one please help me
Question1
ReplyDeletegiven the displacement (s) =8t-t^2
v = d/t
we are given s and v
v = 4
we can differentiate the equation s = 8t-t^2
ds/dt= 8 -2t
since ds/dt = velocity = 4
we can solve for the time
4 = 8 -2t
t = -4/-2
t = 2
Number4
ReplyDeleteE=k/r^2
E=kr^-2
DE/DR
E=2kr^-3
Question10
ReplyDeletey = 7sin x + ln (4x^2 +1)
sub m = (4x^2 +1)
dm/dx = 8x
y = ln m
dy/dm = 1/m
dy/dx = dy/dm x dm/dx
= 1/m x 8x
= (1/4x^2 +1)x 8x
= 8x/4x^2 +1
dy/dx = 7cos x + (8x/4x^2 +1)
Very`helpful
ReplyDeletethax:)
s = 8t -t^2
ReplyDeletev = 4
ds/dt = 8 - (2 x t)
ds/dt = 8 - 2t
to find t we use
4 = 8 - 2t
2t = 8 - 4
2t = 4
t = 4/2
t = 2
s = 20t -2t^2
ReplyDeletewhen brakes is applied it comes to 0.
ds/dt = 20 - (2 x 2)t
ds/dt = 20 - 4t
0 = 20 - 4t
4t = 20 + 0
4t = 20
t = 20/4
t = 5
It takes 5 seconds for the train to stop
replace in equation
s = 20t -2t^2
s = 20 (5) - 2 (5) ^2
s = 100 - 50
s = 50
V = 50(60-t)^2.
ReplyDeletelet m = (60 - t)
Rewrite dv = 50^2
dv/dm = (2 X 50)
dv/dt = dm/dt x dV/dm
When t = 4
dv/dt = 100(60-4)
dv/dt= 100 x 56
64/dt = 5600
I'm not sure if this is correct
E=k/r^2
ReplyDeleteMiss said no downstairs.
E= kr^-2
E = -2kr
h = 550t - 4.9 t^2.
ReplyDeleteh = 550 - (2 x 4.9) t
h = 550 - 9.8 (t)
0 = 550 - 9.8t
9.8t = 550
t = 550/9.8
t = 56.12
Question 9
ReplyDeletey=240(1-e^-.05t)
dy/dt=0-240e^-.05t
letu=-.05
dy/dt=-.05
y=-240e^u
=240e^-.05t
Question7
ReplyDeletey=1.85sin36∏t
dy/dt=1.85cos36∏t
v=1.85cos36∏t
whent=0.025,v=1.85cos36∏(0.025)
=0.1175ms-1
Question6
ReplyDeleten=550t-4.9t^2
dn/dt=550-4.9(2)t^2-1
n=550-9.8t
n=0
0=550-9.8t
-550/-9.8=t
t=56.12
hence the maximum altitude of the rocket attains is substitute t=56.12
h=550(56.12)-4.9(56.12)^2
=30866-15432.33
=15433.7m
s=20t-2t^2
ReplyDeleteds/dt=20-4t
v=20t-4t
v=0
0=20-4t
-20=-4t
-20/-4=t
t=5
substitute t=5 into
s=20(5)-2(5)^2
=100-50
s=50m
Question3
ReplyDeleteletu=50 letv=(60-t^2)
dy/dt=0 dv/dt=2(60-t)^2-1
=-2(60-t)
using the product Rule
dv/dt=udv/dt+vdu/dt
=-100(60-6)
at t =4
v=-100*56
-5600ms-1
1.ds/dt=8t-t^2 =8-2t
ReplyDelete@4=8-2*4=0
2.s=20t-2t^2
ds/dt=20-4t
3.v=50(60-t)^2
v=50(60-t)(60-t)
v=50(3600-60t-60t+t^2)
dy/dx=50(-120t+t^2) = dy/dx=6000t+50t^2=6000+100t
4.e=k/r^2 =kr^2-
de/dr=-2kr^3-
5,v=l(d^2q/dt^2)
dv/dx=l(2qd^2q1- * -2dt)
dv/dt=l2qd^2q-1 * l-2dt
6.h=550t-4.9t^2
dh/dt=550-9.8t
7.y=1.85sin36πt
letx=36πt
dy/dx=113.04t=113.04
dy/dxsinx=cosx
dy/dx=cos36πt*113.04
dy/dx1.85=0
now dy/dt=cos36πt*113.04 (t=.025)
dy/dt=0.75
8,q=ce(1-e^-t/rc)
dy/dx=-t/rc=x
=-1*-1
9.y=240(1-e^-0.5t)
letx=-0.5t
dy/dx=-0.5
dy/dxe^x=e^x=e^-0.5t
dy/dx1-e=1-e^-0.5t
10.y=7sinx+ln(4x^2+1)
letx=4x^2+1
dy/dx=8x
dy/dxln=1/x
dy/dx=1/4x^2+1 *8x
dy/dx7sinx=7cosx
dy/dx=7cosx+1/4x^2+1 *8x
ques..1...
ReplyDeleteThe 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?
s= 8t+t^2
v=d/t
v=4
differentiate the equation
s=8t-t^2
ds/dt=8-2t
since ds/dt=v=4
therefore slove for t
sloveing for t
4=8-2t
t=-4/-2
=2
question 6.
ReplyDeleteThe 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?
h= 550t - 4.9 t^2
dh/dt=550 + 9.8^2
h=the altitude
max altitude=
0=550+9.8t
9.8t=550
t=550/9.8
=56.122
ques 5.
ReplyDeleteok well you have to diff. q = (2t + 1)^.5 -1 twice.
substitute
m = 2t+1
dm/dt = 2t
then rewrite q= m^.5 -1
work out the numbers.Therefore:
q = m^-0.5
dq/dm = -0.5m^-1.5
then :
dq/dt = dm/dt X dq/dm
dq/dt = 2t(-0.52t+1)^-1.5
NOTE: dont forget to replace the substitution made.
Ok... I can't seem to diff this thin thin a second time. Can someone plz help me.
hey computation you didn't finish #6.
ReplyDeleteyou were goin correct, so im just goin to continue from where you left off.
you found t=56.122
the question asked for the altitude h. So to finish the question you have to substitute t=50.122 into the given equation because that is the equation to give you the altitude.
substituting t=56.122 into :
h = 550t - 4.9 t^2
h = 550(56.122) - 4.9(56.122)^2
h = 30867.1 - 15433.4 = 15433.6 => ans