v.s
Differentiate with respect to t.
y = b cos t + t2 sin t
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Shriya
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sin t (2-b) + 2t cos t
v.s
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- sint
Chlorophyll
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( cost)' ? (sint)' ?
Chlorophyll
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-> b ( cos t ) ' = - bsint
Okie, so far?
v.s
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okayy
v.s
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y' = -bsin+2tcos
Shriya
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y=bcost+t2sint
differentiating w.r.t t,
dy/dt= -bsint+ d/dt(2tsint)
now, we have to differentiate 2tsint using product rule for derivative.
so, the derivative of 2tsint becomes (2tcost+2tsint)
therefore, dy/dt= -bsint+2tsint+2tcost
taking sint common,
dy/dx=sint(2-b)+2tcost
v.s
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i got it wrong
v.s
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but i understand what you did
Shriya
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datz ok! :) not a big deal. derivatives is realy easy as compared to the other sections of calculus. u jus need 2 know the ways to differentiate! :)
v.s
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yeaaa thank you :)
Chlorophyll
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Apply product rule here ( t²* sin t ) ' with :
u = t² --> u' = ...
v = sint --> v' = ....
Chlorophyll
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@v.s Do you know product rule?
v.s
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yes
v.s
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2t*sint + cost*t^2
Chlorophyll
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Yep, now put the 2 results together :)
y' = ....
Chlorophyll
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y = b cos t + t² sin t
->y' = ...
v.s
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y"= -bcost+2t*sint+cost*t^2
Chlorophyll
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y = b cos t + t² sin t
-> b ( cos t ) ' = - bsint
-> ( t² sin t )' = 2t sint+ t² cost
v.s
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you just put them together
v.s
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- bsint + 2t sint+ t² cost
Chlorophyll
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What's my instruction above?
v.s
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product rule?
Chlorophyll
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The product rule applied to take derivative of the second term!
v.s
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i'm confused
Chlorophyll
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y = b cos t + t² sin t
-> b ( cos t ) ' = - bsint ( derivative of the first term)
-> ( t² sin t )' = 2t sint+ t² cost ( derivative of the second term)
v.s
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what do you do next?
Chlorophyll
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Yep, now put the 2 results together :)
y = b cos t + t² sin t
y' = - bsint + 2t sint+ t² cost
Chlorophyll
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Now you can make it neat by simplifying - bsint + 2t sint together.
v.s
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okaay
thnks you