Question 1
Water is being drained from a pond such that the volume V (in m^3) of water in the pond after t hours is given by V = 5000(60 - t)^2. Find the rate at which the pond is being drained after 4 h.
Question 2
The velocity of an object moving with constant acceleration can be found from the equation v = (v[0] ^2 + 2as)^2, where v[0] is the initial velocity, a is the acceleration, and s is the distance traveled. Find dv/ds.
Example 3
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.
Example 4
The distance s (in m) traveled 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?
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140 comments:
in example 3 dE/dr
so an expression for this will look like k=r^2E??
for question 1 i tink u have to find dv/dt and subst 4 into dv/dt to find rate. the word "rate" confuses me. does it mean gradiant?
crusier!!! i agree with u, but why u make k the subject of the formula. if u look at the question the key word is "instantaneous rate of change". dis means to differentiate..dE/dr. wat u all tink??
for example 4, does the word stop mean your value of t is 0? i really dont know!!!!
Any suggestions on question 2?? i tink the best method to find dv/ds is by the chain rule, also called function of a function rule.
for exampledy/dx of the function :y=(2x^2 + 2x )^2 can be expressed as 2(2x^2 + 2x)multiply by 4x + 2, which is the differential in the bracket. Anybody ever hear about this method??
Question1 the rate of change would be dv/dt of the equation and subst 4 in the equation.Therefore the equation would be dv/dt=2[5000{60-(4)}]
I am not sure why crusier made k subject of the equation but i think it should be E=k(r^-2),
dE/dr=2kr^-3.
well for question one,i agree with cokebaby, i think that you must first square or expand the numbers that is in the brackets, and then multiply by 5000. then you can use dV/dt, and substitute the value of t= 4 in to the equation obtained.
yeah that true, cruiser, they want the instantaneous rate of change. so i think that you must differentiate that equation, which will give dE/dr= -2kr^-3. correct if i am wrong please.
for example 4, i think that you must use the ds/dt for the equation.if it coming to a stop, doesn't that mean that the differential would equal to 0?
following the procedure of solving the question. first you must rewrite this question so it resembles the equation for gradient dy/dx. once this is done it is quite simple to recognise if the question is a linear equation or according to the situation a quadratic equation. then you solve for the unknown in the equation. you then plug the value into the the equation and then WalaH! you have the answer.Correct!?
Solve this question,
A computer is infected with the Sasser virus. Assume that it infects 20 other computers within 5 minutes; and that these PCs and servers each infect 20 more machines within another 5 minutes, etc. How long until 100 million computers are infected?
-----------------
EQUATION:
N(t) = 20*5^(x/5), where x is number of elapsed minutes.
100x10^6 = 20*5^(x/5)
5x10^6 = 5^(x/5)
Take the log of both sides to get:
log5 + 6 = (x/5)log5
x/5 = 1 + 6/log5
x = 5 + 30/log5
x = 47.92 minutes
2)
f(t)= 10^t," where t is the
elapsed time in minutes. How would you apply logarihm to determine when the sample will grow to 5 billion viruses?
----------------
5x10^9 = 10^t
Take the log of both sides to get:
9 + log5 = t
t = 9.69 minutes
can some shead some light as to exactly how this works out.
but sleepy if u put the differential equal to zero and solve, u wud be finding for "t" the question didnt ask u to find for "t".. they ask u to find "s" i tink dis question missing sumting...
to cruiser's first comment:
the exprsson will be dE/dr. since you are changingelectric field with respect to the distance r you need to change the equation to E=k*r^(-2). this will make the question similar to a normal differentiation.
wel in question 1... t is given as 4 hours. after differentiating dV/dt, substitute the value for t and you will arrive at the answer. the answer will be in units of m^3/h.
some help with question two. i'm seeing too many unknowns and not enough information....
to sleepy:
in example four i believe your statements are accurate. you are finding ds/dt as you stated. this will give you the velocity of the train at any one point in time. when you put this eual to zero it signifies that the velocity has become 0 therefore the train has stopped.
ichigo is right. why did cruiser change the subject of the formula in quest. 3. if the question asks for the rate of change of the electric field with respect to r, then the differential is
dE/dr
crimson crock!!! does dat procedure apply for example 4?? so u tryn to say i cud differentiate s = 20t - 2t^2...put the differential equal to 0...then solve for t and then subst t back into the original equation to get s.. dats wat ur sayn?? Im confused!!!
to wong fei:
in question two a = v/s which can be changed to the form v = vs^-1, this should get rid of unwanted unknowns
queat 4... a value for t is missing??? should there be a given value?
hellion: makes sense
for question one, does anyone know how to difffrentiate the equation without expanding it?
to wong fei hong:
in question 4 T is 4 hours
wong fei hong!!! all dat question 2ask u for is dv/ds. so just differentiate the function with respect to s..its dat simple!!!
are you sure hellion. in quest 2, t = 4
not quest 4???
verify this for me please
word cokebaby..ah now see that
hellion!!!! of course u can differentiate a function without expanding it. did u read my fourth comment??? function of a function rule!!!! consider question 1... dv/dt wud be 10000(60-t) multiply by the differential in the bracket which is -1...u wud get -10000(60-t...u understand????
cokebaby is absolutely right. well said
for question 1 you would hav to differentiate the equation. and then substitute t=4 into the equation to the volume.
cokebaby so u can differentiate when u hav brackets by takin d power and * it by d # infront of the brackets and then minusing the power like normal.
Q2 is not clear to me could someone clarify for ne please??????
we could re-arrange the equation (Q3)
E= k(r^-2)
so
dE/dr= -2k(r^-3)
moose question 2 is very simple. just differentiate v = (v[0] ^2 + 2as)^.5, ie find dv/ds read my previous comments..
yes pussinboots!!!!! but dont 4get to multiply by the differential of wats inside the brackets ok
No one has respoded to my comment. what's up with that? As a student who is new to these concepts I need people's input.
dy/dy = to something, exactly what is this expression called.
for this question i think that you have differentiate s with respect to t.
eample 4:
s=20t - 2t^2
ds/dt=20 - 4t
i think that will be the answer to this question because it is the rate of change before it goes to rest. if i'm wrong could someone please correct me. thank you
crimson crock with respect to your last comment the expression
"dy/dy = to something"...its dy/dx and not dy/dy..dy/dx simply refers to the differential of the function y with respect to x. understand???
the time it will take for the train to stop will be the rate of change of s wtih respect to t which is ds/dt
so
ds/dt= 20-4t
to any1!!! with respect to crimson crock how did he get this line...
100x10^6 = 20*5^(x/5). Can sum1 explain to me how did he get dat. Crimson crock, can u explain??? Check his second comment.
will i un derstamd thaat u have to find dv/dt but t=hours so t=4h. first u differentate, and then u substitute 4 in the equation.
eh cokebaby watch the equation N(t) is 100 million which is 100^6
watch it good cokes
Q1. when a question states find the rate, it means some variable over time. so you find dV/dt then substitute in the formula when t = 4h. so dV/ dt = 1000t^1.when t= 4
dV/dt = 4000m^3. someone commment on this.
for question 2
v = (v[0] ^2 + 2as)^.5
dv/ds = 10v^9 + 10as^4
someone correct me if i am wrong.
Q3. since k is a constant, constants cannot be differentiated so k is 0 . dE/ dr = 2r^1
like what puss-in-boots and cokebaby said for e.g.1 you have to the dv/dt of the equation and and the substitute 4 into the equation
in example 2 the equation must first be simplified and the brackets remove before it can be differentiated
CAN SOMEONE EXPLAIN QUESTION 2
in example 3 the equation can be differentiated as is, using dE/dr
it will be the same as E=k*1/r^2 or E=k*r^-2 this would be differentiated to give -2kr^-3
i think in example 3;E=k/r^2, you would first have to rearrange the equation to y= ax^n and then solve.
example 3;
E=k/r^2
E=Kr^-2
dE/dr=-2kr^-3
so therefore i totally agree with lilo
to: apocalypse,
i may not be correct but i think that in example 2 you would have to differentiate as always.
example 2;
v=(v[0]^2 + 2as)^5
dv/ds= v[0]^10 + 2as^5
dv/ds= 10v[0]^9 + 10as^4
please correct me if i'm wrong thanks.
in example one how do you put the equation in the form dv/dt in order to differentiate it.
n2o it is a simple concept where you first determine the relatively changing variable, which in most cases would be seen as x, as a pose to y which is proportionate to x...in q1, u first expand the equation to see how many variables u are working with, then when expanded u differentiate by multiplying the powers or indices of the variables by the coefficients and then minusing 1 away from the indices; if it is already an interger it is taken as 0 or insignificant in the dv/dt expression...after u substitute the value of t in the equation to obtained the rate of change...i hope u understand. i tried to break it down simply as possible
for example 3 the expression would look like
E=K/r^2
E=Kr^-2
E= -2Kr^-3
Apocalypse..U didnt read the previous comments. Why? If u did u wud hav had an idea how to work out Question 1.dV/ dt can neva be 1000t^1. Please read my 4th comment in relation to function of a function rule.
Apocalypse!!! wat u dont understand about Question 2. Again i say if u had read the previous comments u wud have had an idea how to work it. All the question ask for was dv/ds..so just differentiate the function v with respect to s. Again function of a function rule. Read my 4th comment.
do u have to raise the entire expression from question to the power 2 or is it just what is in the brackets
I agree with cokebaby, thats corect as far as im concerned.
well ratman...jus wats inside the bracket wud be raised to the power 2. the firat step in differentiating wud be to multiply the power by the coefficient. Understand???
in exzmple 4
after u differentiate the equation s= 20t - 2t^2
u should get
s=20 -4t
since u have two unknowns how can u solve the equation to get a value for s?
In example 3 it is requested that the rate of change is electric (E) with respect to distance (r). In addition the rate of change is instantainous. All this translates to is dE\dr.
The solution therefore is
E= k\r^2
E= kr^-2
dE/dr= -2kr^-3
in example 4 what does t represent
n20!!!! dat last question missing sumting!!!!wat u tink? how about finding ds/dt, then substituting
t=0 into ds/dt?
Cruiser what is the purpose of making k the subject of the formula
lilo.. in example 4, the "t" represents time.
In question 1 in order to find the rate at which the pond is being drained a relationship between volume and time must be established this relation ship is dv/dt. After dv/dt is established then 4h (time) can be substituted for t and the rate at this time can now be found.
Lilo in example 4 s represents distance travelled and t represents time.
Cokebaby in example 1 the amount of water that is being drained from the pond at a given time is not constant. If it was consant then we could have just simply substituted the value 4 as t and determine the volume of water drained at this time (straight line graph concept). The volume of water being drained at time 2h would be different from the volume of water being drained from time 3h and 4h. The word rate describes how much water (v) is currently being drained at a given time (t). Remember the rate is not constant, this is why we have to find a relationship between volume and time (dv\dt). (curve line graph concept). Remember dv\dt means volume with respect to time so as volume changes time changes as well.
Crimson crock if you mean dy/dx= some expression then it means the following:
y= a variable such as time or distance
x= another variable again such as time or distance. There are many variables that is why you may come across many expressions not just dy/dx such as dr/dt, or de/dz or d(any letter of the alphabet)/d(any letter of the alphabet). The letter after the d simply represents some variable. Coming to your question now dy/dx means what is the relationship between y and x. It may also be translated as what is the relationship between two variables on a curved gaph.
Crimson crock lets look at the dy/dx concept in a more practical sense. lets take example 1. Analysing it some pond is being drained. The rate at which the pond is being drained is not constant (keep this in mind). because of this a relationship between the volume and time must be found.
y is now v (volume)
x is now t (time)
so it becomes dv/dt (volume with respect to time)
i agree with variable 47. the first equation is E = k/r^2, therefore the instantaneous rate of change would first be dE/dr= -2kr^-3 which is really
dE/dr= -2k/r^3.
to everyone!!! take some practice...
y = -2x[-3] + 4x[-2]. Find dy/dx
variable 47 you said in 1 of your comments that the rate at which the water is being drained is not constant, why did you say that? when you find dv/dt that will be like the rate of change which to me is like a constant.
eh somebody help me with q2 na please?????????????????????????????????????????????????????????????????????????
i am not too sure about example 4 but i think that if s = 20t-2t^2
ds/dt= 20- 4t^1
the gradient which is ds/dt =0 then the distance travelled would work out to be 20-4(0) which is 20metres ... please let me no if i am correct...
y=-2x[-3] + 4x[-2]
dy/dx = 6x[-4] + (-8x[-3])
dy/dx = 6x[-4] - 8x[-3]
correct me if i'm wrong
thanks for the practice cokebaby. i think the answer is
dy/dx=6x^-4 - 8x^-3. am i right?
to cokebaby
dy/dx= 6x[-4]-8x[-3]
about my answer, the numbers inside the brackets are the powers.
in example 1, differentiate V with respect to t i.e. find dV/dt. that will be:
dV/dt = -10000(60-t)
then substitute t=4 in dV/dt.
to answer cokebaby's question, the word rate is used when a variable changes with respect to time.
please correct me if i'm wrong!!!
y = -2x[-3] + 4x[-2]. well lets see.
dy/dx = 6x[-4] - 8x[-3]
poison think about it. Let us assume it was a constant then could we just substitute 4h into the equation and find the value for volume (v)? Remember we found dv/dt in the first place because the rate of volume is continuously changing.
eh cokebaby i see you telling people bout q2 work it out for me please thnaks
variable 47 are you saying that differention is to calculate the rate of change of a gradient?
people solve this;
y=4x^2 - 1/2^3 + x . what is dy/dx?
yes you can say that but do not misinterpret the concept.
in example 2 you differentiate V with respect to s. therefore,
dV/ds= 10V[0]+10a (V[0]^2+2as)^4.
the method i used to find dV/ds is the same as cokebaby's chain rule.yes cokebaby! i know of this method
To Poison:
Why should we differentiate a function?
When do you think we should differentiate?
After examining these comments I finally understand what I need do from question3 = EKr^-2
=E=-2Kr^-3 by solving in dE/dr
i believe that the answer you submitted sparkle is very accurate. the ds/dt= 20-4t^1 and since it is coming to a stop, t would =0 so the answer sparkle had acquired, i suppose is correct. (20m)
Cruiser answer this why is it dE/dr and not dr/dE?
moose for question 2! u wud have to expand out the brackets. so u wud get 2as[2]. so the differential dv/ds is 4a.. i tink.
sleepy, the answer to your question is 8x + 1. Thank sleepy, now ah feelin bright
in example 3 the instantaneous rate of change of the electric field with respect to r i.e.
dE/dr = -2kr^-3
so cokebaby yuh saying that
v(o)^4+2as^2
dy/dx= 4v^3+4a
this is it??????????
you differentiate a function when you want to find the gradient or rate of change and you differentiate only when you know the gradient is not constant at all points like in a curve in a graph.
issippii think that u differentiate w.r.t. t. so ds/ dt = 20 - 4t^1. according to the question 'when coming to a stop', the velocity is zero, so ds/ dt is zero, because rate of change of distance over time.
then u find t when v = 0, so t= 5. but wat i dont understand, when u find the distance , usung the normal formula, the distance is 0, cause v is 0. can someone explain as soon as possible. my patience is getting thin!!!!!!!!!!
i think to do number two, you should approach the problem like a normal diffrentiation sum. Don't get afraid when you see dv/ds. It is the same concept as dy/dx. It is just the rate of change of the velocity with respect to the distance travelled. Just diffrentiate the equation and you would arrive at an expression for dv/ds.
i agree with boo boo baby because in sleepy's question the middle term has no x in it so therefore you would just have differentiate
4x^2 + x which will give 8x + 1
the rate at which the pond would be drained can be found by dv/dt. in which the value 4 can be substituted.2[5000(60-{4}]
in eg.3 the instantaneous rate of change of the electric field with respect to r can look like this;E=k(r`2) ;de/dr= -2kr`-3.
ok for example,3 u have to find dE/dr . now u take out the constant,K and then u are left with E=r^2
now we differentiate E=2r.
But I have a question when I take out the constant,k the sum said divide by r^2 and i ignored that, is that wrong??????????????????????????????????????????????????????????????????????????????????????????????????????
Orrrr! i now understand Q4. so when we differentiate s with respect to t we get velocity, and the final velocity is zero so we have to put ds/dt = 0 and solve for t. then take the value the value of t and substitute it into the original equation s = 20t - 2t^2. Deres ur answer.
crimson crock, you really ask some challenging questions. well i think the function would be:
N(t)= 20*5^(x/5), i dont understand how you arrivd at that.
Again example three is just like a normal deffrentiation sum. However you must first convert the equationinto a form where it can be diffrentiated. So, what you have to do is bring r^2 with the numerator so you would result with a equation E = kr^-2. From this form, we can cane treat this as a normal diffrentiation sum resulting in an equation dE/dr = -2kr^-3. Please correct me if i am wrong.
for example 4 i believe that you should go about saying ds/dt = 20-4t and and after this i am a little lost cud some one assist me
for question 2 in determining the expression in terms of dv/ds e could say by solving and deffrentiating since v0 is constant
i have a question..if an ice cube is melting uniformly, wat is the expression for the instantaneous rate of change of surface area (A) of the cube with respect to the edge (e). Is dA/de =12e or
dA/de = -12e. Wat the answer and why..??????
I THE MIGHTY APOCALYPSE HAVE A QUESTION!!!!!!!!!
Find the derative of a function f: => x^3 - 1( where x is being real) at x. what is the gradient of the curve y =x^3 - 1 when x = 2? find the equations of the tangent and normal to the cuve y = x^3 - 1 at the points (2,9).
HE WHO ANSWERS THIS QUESTION IS CALLED THE MATHEMATICIAN OF ALL AGES!!!!!!!!!!!!!!!!!!!!!!!!!!!!
HAS NO ONE ANSWERED MY QUSTION? PETTY HUMANS!
HAHAHAHAHAHAHAHAHAHAHA!!!!!!!!
Apolcalypse!!!! dats very easy!!!!!! dy/dx = 3x^2. the gradiant at x =
2 wud be 12.Just substitute
The equation of the tangent and the normal to this curve y=mx+c, subst values into the equation
9 = 12(2) + c.. so c is equal to
-15, so the equation of the tangent is y = 12x - 15. The gradiant of the normal wud be -1/12 so the equation of the normal is Y = mx + c which is
9 = (-1/12)(2) + c. C wud be equal to 55/6. so the equation of the normal is y = -1/12 + 55/6..Apocalpyse!!! please dont bring so lengthy question nex time...
cokebaby. i dont beleive it! this.........answer u gave me is .............is.......................W.R.O.N.G,INVALID, NOT RIGHT,WRONG LIKE A CIRCLE NOT CORRECT!!!!!!!!!!!!
PUNY HUMAN! ONCE AGAIN NO ONE (BUT ME) IS SUPERIOR!!
BUT ACCORDING TO THE RULES i will have to help u.for the first question ,your answer was right. for the 2nd, check ur signs, for trhe 3rd, check ur intercept!
NO LONG QUESTIONS EH!
FOR THAT I WILL NOT PUT 1 LONG QUESTION BUT 2 QUESTIONS!!!!!!!!!!
HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA!UNTIL NEXT TIME COKEBABY. CHECK UR ANAWERS.
why does sparkle even bother to com on the blog, his/her comments dont even make any sense. i think he/she needs the most help out of all of us
i agree with cokebaby for the first answer, but the second answer would be y = 12x + 15 and well for the normal, i think cokebaby put the right answer. what is the right answer apocalypse?
the great apocalypse has another question.
for what values of x is the tangent to the curve of y = 3x^2 - 6x parallel to the x axis(that is where the slope is zero)
ok, ok i beleive i was too harsh with this human life form called cokebaby, at least he TRIED.good attempt.the answers are
1. 12
2. y = 12x + 15
wait.........analysing cokebaby's question again................WHAT!!!!!.............HOW CAN THIS BE.........I.I..I.....I can't believe i made a technical error..........how...can .....this .......be! for the third question..........that.............is corect.
my mistake cokebaby. i must comment on sleepy for being alert. like u are not sleepy after all!!!!! but i apologise for my mistakes, living with u humans have made me weak!!
THE FIGHT IS NOT OVER COKEBABY, IT HAS NOW BEGUN!! ILL BE WATCHIN U TOO SLEEPY!!
UNTILL NEXT TIME. LET THE FORCE BE WITH U
APOCALPYSE I don't know which galaxy you come from but here on earth when we get a mathematics exam we get 0 for putting answers alone like what you did in your last comment(February 26, 2008 4:50 PM)
Take my human based advice: SHOW YOUR WORKING!!
APOCALYPSE I can work out the first part of your first challenge:
It is simply to find the first derivitive of the function:
let f(x) = y
y = x^3-1
dy\dx = 3x^2
by asking what is the gradient of the function when x=2 is the same as asking: what is dy\dx when x=2 because dy\dx in this case is a gradient function therefore all we have to do is substitute x=2
=> dy\dx = 3x^2
substituting x=2
dy\dx = 3(2)^2
dy\dx = 3(4)
dy\dx = 12
You see STEPS this is what we earth people do.
APOCALYPSE I can't quite understand the second question of your first challenge but i see that you recently posted an answer for it. Since you are so great could you please explain how you got the solution y=12x+15. Thank you your greatness.
Cokebaby that is a very nice question. I think the solution to your answer would be dA/de = -12e
Reason being is that since the ice cube is melting then the gradient would be decreasing and likwise if it was reresented on a graph then the graph woud have to depict a nagative gradient direction. Is this correct?
Not one of you dudes even tried to answer the question i posted earlier. Come on what's up with that!!!?
forquestion one i believe you need to expand what is written in brackets then multiply by 5000. After getting your equation then you differentiate. You will get dy/dt= 18000000-600000t+5000t^2. you then substitute the value given for t.
for question 2 i saw someone wrote something about chain rule.please enlighten me further.thanx a bunch.
In question 3 E=k/r2
E= kr^-2
dE/dr=-2kr^-3 is this expression correct!!!!!
Q4. s=20t-2t^2
ds/dt=20-4t
Im not sure im on the right track but i think u put this equal to zero, then u have 20-4t=0
So 20 = 4t
t=5 metres
??????????????
What do you guys think
in example 4 how would you determine the distance
To me I would say that s=20t-2t^2 could we say that we can find derive these examples by stating
4t^1????
cruiser. did u read all the comments? i asked that question already and i believe cokebaby and moose answered that question.
read over the comments
hmmmmmmm. VARIABLE 47!!!!!!!!
do u knw who i am?DO YOU?
as u humans refer yourselves as homo sapiens, i am homo superior. im am a child of the atom . i am an X MAN! one of the first too!my real name is En Sabah Nur.i am an immortal bonded with extraterrestrial technology.
since u ask to explain the question, i will but not now. i time travel on thursdays. see u next 200 years in the furture!
I agree with cokebaby.The rate of change would be dv/dt of the equation and substitute 4 in the equation.
Wow crimson crock, where did you get those lovely questios, quite interesting.
in example 3 the instantaneous rate of change means that you want to find dE/dr.Also the differnciate of dE/dr= (2-)(k)r ^-3
For Question 1 I agree with Coke Baby and Sleepy (60-t)^2 should be expanded out first then each term multiplied by 5000, the expression for dV/dt found and the rate (dV/dt) calculated by sub-ing t=4 in the equation.
For Question 2 when the equation is expanded it would be (v[o]^2)^2 + (2as)^2 =(v[o])^4 + 4(a)^2(s)^2
So I think you have to deal with the variables v and s only and a be treated as a constant.
Therefore:
dv/ds=(4v[o])^3 + 8(a)^2(S)
For Question 3
Well this can be rewritten as :
E = k x (r)^-2
Then dE/dr would be the expression for the instantaneous rate of change of the electric field with respect to r.
Therefore:
dE/dr= -2k(r)^-3 or -2k/(r)^3
Hey I'm Not sure for Question 4 but I think that after finding ds/dt and equating to 0 The time for the train to stop will be found and then if this time could be substituted back into the original equation the distance it takes to stop will be found.
for question 1 I believe that the question ask for the rate of dv/dt; after which t=4 can be simply substituted into the formula.
this can be found by breaking up the polynomial & find dv/ds by the differentiation of the seperate terms
to solve the expression for the rate of change with respect to r = constant = Er^2
this can be solved similarly to the question 2.
for qustion 1 i agree with cokebaby that u have to find dv/dt and subst 4 into dv/dt to find rate, for cokebaby question if rate is the same as gradient, i think that the relationship there is if u plot the points on a graph for the rate, it would be equal to the gradient... am not sure but its wat i think..
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