Question 1: A curve has equation y = 4/(2)^.5, find dy/dx
Well i dont know how to start this question since there is no x variable. Since this is so i believe y cannot find y with respect to x which is dy/dx..
Question 2 A curve is such that dy/dx = 16/x^3 and (1,4) is a point on the curve, find the equation of the curve.
Well to find this we have to find the intergral od dy/dx with respect to x. Am not sure what the points (1,4) has to do with the question, but i believe that it is the upper and lower limit of the intergral of dy/dx.......
actually "swanky thing" i think that answer is wrong because the -sin 2 thetha is a linked term therefore u have to differentiate the terms differently then put them back together
therefore
lets make theta " r" to make life easier ok i mean for the whole question ( y= 6r- sin 2r)
then to differentiate -sin 2r
let m = 2r dm/dr = 2
and let t= - sin m then dt/dm = -cos 2r ( when m is subs. back into the equ.)
thus dt/ dr = dt/dm X dm/dr = - cos 2r X 2 = - 2 cos 2r
with respect to the first question i believe that miss was trying to put x where the 2 is. so based on that the ques wud be stated like this
y = 4/(x)^5
and the question is sright forward ....jus bring all the terms from downstairs to upstairs and then find dy/dx jus make sure when doing this u dont mix up the signs when multipling and u'll get the required answer
ok swanky thing WRT to ques 2 it is alot simplier than it looks. all u have tuh do is integrate the equation then you will have to use the two pts that were given (1,4) to find for C
so u plug it in to the eqa and the value u get will be C then you jus rewrite the eqa and thats the answer
just make sure when u are working out your question to rush it down because u make a small mistake in your ques you forgot to differenate the 6r term so your answer suspose to be
Ques2. A curve is such that dy/dx = 16/x^3 and (1,4) is a point on the curve, find the equation of the curve i think i am finally getting this deff thing by reading what renjisan said.
Ques3. y = 6theta - sin 2 theta, find dy/dx well this would be deff with respect to theta and the integral of sin is cos so the ans should be 6-cos2theta
ok animechic it depends on how the question is stated unless they ask u for the equ of the curve you dont have to state it since it is just differentation that is the equ of the curve someone correct me if am wrong
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Question 1:
ReplyDeleteA curve has equation y = 4/(2)^.5, find dy/dx
Well i dont know how to start this question since there is no x variable. Since this is so i believe y cannot find y with respect to x which is dy/dx..
Question 2
ReplyDeleteA curve is such that dy/dx = 16/x^3 and (1,4) is a point on the curve, find the equation of the curve.
Well to find this we have to find the intergral od dy/dx with respect to x. Am not sure what the points (1,4) has to do with the question, but i believe that it is the upper and lower limit of the intergral of dy/dx.......
Can someone help me out here.
Question 3
ReplyDeletey = 6theta - sin 2 theta, find dy/dx.
I believe miss made a mistake here, is it dy/d theta we are looking for.
dy/d theta = 6 - cos 2 theta
if there is no x, you not suppose to use the substition thing?
ReplyDeleteactually "swanky thing" i think that answer is wrong because the -sin 2 thetha is a linked term therefore u have to differentiate the terms differently then put them back together
ReplyDeletetherefore
lets make theta " r" to make life easier ok i mean for the whole question ( y= 6r- sin 2r)
then to differentiate -sin 2r
let m = 2r
dm/dr = 2
and let t= - sin m
then dt/dm = -cos 2r ( when m is subs. back into the equ.)
thus dt/ dr = dt/dm X dm/dr
= - cos 2r X 2
= - 2 cos 2r
putting all together now
y = 6r - sin 2r
dy/dr = 6 - 2 cos 2r
At least that is how i remember to work it !!
lets make theta " r" to make life easier ok i mean for the whole question ( y= 6r- sin 2r)
ReplyDeletethen to differentiate -sin 2r
let m = 2r
dm/dr = 2
and let t= - sin m
then dt/dm = -cos 2r ( when m is subs. back into the equ.)
thus dt/ dr = dt/dm X dm/dr
= - cos 2r X 2
= - 2 cos 2r
Well i get where your coming from "johnsmith".
ReplyDeleteThanks for the correction/
with respect to the first question i believe that miss was trying to put x where the 2 is. so based on that the ques wud be stated like this
ReplyDeletey = 4/(x)^5
and the question is sright forward ....jus bring all the terms from downstairs to upstairs and then find dy/dx jus make sure when doing this u dont mix up the signs when multipling and u'll get the required answer
y = -20/x^6
QUES2
ReplyDeleteok swanky thing WRT to ques 2 it is alot simplier than it looks. all u have tuh do is integrate the equation then you will have to use the two pts that were given (1,4) to find for C
so u plug it in to the eqa and the value u get will be C
then you jus rewrite the eqa and thats the answer
y = -8/x^2 + 12
FLASH
ReplyDeletejust make sure when u are working out your question to rush it down because u make a small mistake in your ques you forgot to differenate the 6r term so your answer suspose to be
= 6 - 2cos2teta
ques 3
ReplyDelete6theta-sin2theta
firstly you look to see what trig function is used.
so here we have sin, and the integral od sin is cos .
so therefore the ans will be
dy/dtheta= 6- cos 2theta
ques 3
ReplyDeletefirst you look to see what trig function is used.
we know the integral of sin is cos.
so the ans = 6 -cos2theta
Ques1.A curve has equation y = 4/(2)^.5, find dy/dx
ReplyDeletei don't understand this question
Ques2. A curve is such that dy/dx = 16/x^3 and (1,4) is a point on the curve, find the equation of the curve
ReplyDeletei think i am finally getting this deff thing by reading what renjisan said.
Ques3. y = 6theta - sin 2 theta, find dy/dx
ReplyDeletewell this would be deff with respect to theta and the integral of sin is cos so the ans should be 6-cos2theta
dutchess can u breakdown question 3 and how do end up get the equation 4 d curve
ReplyDeleteQuestion2
ReplyDeletedy/dx=16/x^3
=16x^-3+c
=16/-3+1 x^-3+1+c
16 x^-2+c/-2
y=-8/x^2+c
4=-8/1^2+c
4=-8+c
4+8=c
c=12
y=-8/x^2+12
ok animechic it depends on how the question is stated unless they ask u for the equ of the curve you dont have to state it since it is just differentation that is the equ of the curve someone correct me if am wrong
ReplyDeleteSo "renijan" so i intergrate question 2 and just substitue the values for x and y in the equation to find for C...
ReplyDeleteA curve is such that dy/dx = 16/x^3 and (1,4) is a point on the curve, find the equation of the curve.
ReplyDeletedy/dx = 16/x^3
make simlier
dy/dx = 16x^-3
y = 16x^-2/-2 + C
Make the equation simplier
y = -8x^2 + C
points (1,4), where x=1 y=4
4 = -8(1)^2 + C
C = -8(1)^2 - 4
C = 64 - 4
C = 60
Therefore the equation now looks like
y = -8x^2 + 60
I tried so someone correct me if am wrong
Swanky Thing... WRT ques 2. you made simple mistakes, and you really didn't really check yuor police...lol
ReplyDeleteWell you integrated correct. after that you started making mistakes, So i'll continue from there.
when you integrated you got:
y = 16x^-2/-2 + C
you made your first mistake here. simplifying.
y = -8x^-2 + c
then you substitute the point given (1,4) into this equation.
4 = -8(1)^-2 + c
4 = -8 + c
4 + 8 = c = 12
rewrite the equation:
y = -8x^-2 + 12 => equation of the curve.
johnsmith for #3 you are quite correct. those other guys should check there ans.
ReplyDelete