Core math help :(
The parametric equations of a curve are
x=acos^3t , y=asin^3t
where a is a positive constant and 0< t <half pi
i.) Express dy/dx in terms of t
ii.) show that the equation of the tangent to the curve at the point with parameter t is: xsint + ycost = asint.cost
iii.) hence show that,if this tangent meets the x-axiz at X and the y-axis at Y, then the length of XY is always equal to a.
Ive managed to solve i.) which i got as being 3asin^2t.cost/-3acos^2t.sint... giving
dy/dx= sint/cost
from there on im clueless..ive tried but im just -__-
please help
x=acos^3t , y=asin^3t
where a is a positive constant and 0< t <half pi
i.) Express dy/dx in terms of t
ii.) show that the equation of the tangent to the curve at the point with parameter t is: xsint + ycost = asint.cost
iii.) hence show that,if this tangent meets the x-axiz at X and the y-axis at Y, then the length of XY is always equal to a.
Ive managed to solve i.) which i got as being 3asin^2t.cost/-3acos^2t.sint... giving
dy/dx= sint/cost
from there on im clueless..ive tried but im just -__-
please help
by acos^3t i assume you mean a(cost)^3?
(edited 13 years ago)
Original post by connor ellis
by acos^3t i assume you mean a(cost)^3?
yh... basically a.cost.cost.cost
For part i use the formula dy/dx = (dy/dt)/(dx/dt)
whats the answer? i just did it in my head quickly, got -tant
i) dy/dy= -tan t = -sin t/cos t
ii) use y - y1 = m(x - x1) formula, where m is your dy/dx, y1 and x1 (subscript 1s) as what they have given you...
so you should get y - y1 = (-sin t/cos t)(x - x1). expand and rearrange to give: xsin t + ycos t = y1cos t + x1sin t
sub y1 = a(sin^3)t, x1 = a(cos^3)t
to give: xsin t + ycos t = a(sin^3)t.cos t + a(cos^3)t.sin t
factorise RHS: xsin t + ycos t = asin t.cos t[(sin^2)t + (cos^2)t] , remember that (sin^2)t + (cos^2)t = 1 by using identities
so in the end you will get: xsin t + ycos t = asin t.cos t
ii) use y - y1 = m(x - x1) formula, where m is your dy/dx, y1 and x1 (subscript 1s) as what they have given you...
so you should get y - y1 = (-sin t/cos t)(x - x1). expand and rearrange to give: xsin t + ycos t = y1cos t + x1sin t
sub y1 = a(sin^3)t, x1 = a(cos^3)t
to give: xsin t + ycos t = a(sin^3)t.cos t + a(cos^3)t.sin t
factorise RHS: xsin t + ycos t = asin t.cos t[(sin^2)t + (cos^2)t] , remember that (sin^2)t + (cos^2)t = 1 by using identities
so in the end you will get: xsin t + ycos t = asin t.cos t
Original post by PCL11
i) dy/dy= -tan t = -sin t/cos t
ii) use y - y1 = m(x - x1) formula, where m is your dy/dx, y1 and x1 (subscript 1s) as what they have given you...
so you should get y - y1 = (-sin t/cos t)(x - x1). expand and rearrange to give: xsin t + ycos t = y1cos t + x1sin t
sub y1 = a(sin^3)t, x1 = a(cos^3)t
to give: xsin t + ycos t = a(sin^3)t.cos t + a(cos^3)t.sin t
factorise RHS: xsin t + ycos t = asin t.cos t[(sin^2)t + (cos^2)t] , remember that (sin^2)t + (cos^2)t = 1 by using identities
so in the end you will get: xsin t + ycos t = asin t.cos t
ii) use y - y1 = m(x - x1) formula, where m is your dy/dx, y1 and x1 (subscript 1s) as what they have given you...
so you should get y - y1 = (-sin t/cos t)(x - x1). expand and rearrange to give: xsin t + ycos t = y1cos t + x1sin t
sub y1 = a(sin^3)t, x1 = a(cos^3)t
to give: xsin t + ycos t = a(sin^3)t.cos t + a(cos^3)t.sin t
factorise RHS: xsin t + ycos t = asin t.cos t[(sin^2)t + (cos^2)t] , remember that (sin^2)t + (cos^2)t = 1 by using identities
so in the end you will get: xsin t + ycos t = asin t.cos t
Thanks sooo much!!! i finally get it now
Original post by PCL11
i) dy/dy= -tan t = -sin t/cos t
ii) use y - y1 = m(x - x1) formula, where m is your dy/dx, y1 and x1 (subscript 1s) as what they have given you...
so you should get y - y1 = (-sin t/cos t)(x - x1). expand and rearrange to give: xsin t + ycos t = y1cos t + x1sin t
sub y1 = a(sin^3)t, x1 = a(cos^3)t
to give: xsin t + ycos t = a(sin^3)t.cos t + a(cos^3)t.sin t
factorise RHS: xsin t + ycos t = asin t.cos t[(sin^2)t + (cos^2)t] , remember that (sin^2)t + (cos^2)t = 1 by using identities
so in the end you will get: xsin t + ycos t = asin t.cos t
ii) use y - y1 = m(x - x1) formula, where m is your dy/dx, y1 and x1 (subscript 1s) as what they have given you...
so you should get y - y1 = (-sin t/cos t)(x - x1). expand and rearrange to give: xsin t + ycos t = y1cos t + x1sin t
sub y1 = a(sin^3)t, x1 = a(cos^3)t
to give: xsin t + ycos t = a(sin^3)t.cos t + a(cos^3)t.sin t
factorise RHS: xsin t + ycos t = asin t.cos t[(sin^2)t + (cos^2)t] , remember that (sin^2)t + (cos^2)t = 1 by using identities
so in the end you will get: xsin t + ycos t = asin t.cos t
what about iii.)
Original post by connor ellis
For part i use the formula dy/dx = (dy/dt)/(dx/dt)
can u help with the third paart of the question
Original post by PCL11
hmm, iii) is a weird one...
i was thinking that there'd be two sets of coordinates like
(asint,0) for the x axis
(0,acost) for the y axis
how do u then find the length
Original post by CreativeBetch
i was thinking that there'd be two sets of coordinates like
(asint,0) for the x axis
(0,acost) for the y axis
how do u then find the length
(asint,0) for the x axis
(0,acost) for the y axis
how do u then find the length
Pythagoras theorem?
HINT: sin^2(t) +cos^2(t) = 1
Original post by tw7055
Original post by tw7055
Pythagoras theorem?
HINT: sin^2(t) +cos^2(t) = 1
HINT: sin^2(t) +cos^2(t) = 1
ah! makes sense
Original post by tw7055
Pythagoras theorem?
HINT: sin^2(t) +cos^2(t) = 1
HINT: sin^2(t) +cos^2(t) = 1
I guessed that and i know im almost at the answer...i just dont know how to put it in words. Could you help me write it out
$Original post by Dpwalker
At X put y=0 x therefore =A cost (A cost, 0)
At Y x=0 y= a sint (0, a sin t)
XY length= sq rt (a cost)^2+(a sin t)^2=sq rt a^2= a
Did on iPhone so bit rough :P
At Y x=0 y= a sint (0, a sin t)
XY length= sq rt (a cost)^2+(a sin t)^2=sq rt a^2= a
Did on iPhone so bit rough :P
Thank youuuu
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