Calculus 3 Dirivative

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lukasz

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calculus 3 derivative: T(t) = (i+6tj+tk)/sqrt(1+37y^2)

Can someone show me how to find this derivative because i've been trying to finish it and i cannot get the answer

T(t) = (i+6tj+tk)/sqrt(1+37y^2)

N(t)=T'(t)/|T'(t)|=[-37ti+6j+k]/(sqrt(37)*sqrt(1+37t^2))
 
Re: calculus 3 derivative

lukasz said:
Can someone show me how to find this derivative because i've been trying to finish it and i cannot get the answer

T(t) = (i+6tj+tk)/sqrt(1+37y^2)


N(t)=T'(t)/|T'(t)|=[-37ti+6j+k]/(sqrt(37)*sqrt(1+37t^2))
Click to expand...

Please show us your work - and exactly where you are stuck - so that we know where to begin to help you.
 
Re: calculus 3 derivative

i am stuck here
its right after i use the quotient rule (vu'-vu')/v^2

(((1+37t^2)^(1/2)(6j+k))- ((i+6tj+tk)(37t)))/(sqrt(1+37t^2)1+37t^2)

how do you simplify the numerator? because in my notes my teacher somehow extracted a (1+37t^2)^(-1/2) out of the numerator leaving it as (1+37t^2)^1. well i dunno but i am stuck thier.
 
Re: calculus 3 derivative

you can also right this as ((1/(1+37t^2)^(1/2))*(i+6tj+tk) that way you can take the product rule instead :/ i think ??
 
T(t) = (1/sqrt(1+37t^2))*(i+6tj+tk)
What is the T'(t) ? show steps please


N(t)=T'(t)/|T'(t)|=(1/(sqrt(37)*sqrt(1+37t^2)))*[-37ti+6j+k]
 
Re: calculus 3 derivative

T(t) = (i+6tj+k)/(1+37t^2)^(1/2)

Use quotient Rule - vu'-uv'/v^2

T'(t) = (-37ti+6j+k)/(1+37t^2)^(3/2)

Solved it on my own :D
 
T(t)=a(t) b(t)

a(t)= [1+37t^2]^-1/2
b(t)=I+6tJ+tK

dT/dt=a(t)b'(t)+b(t)a'(t)

a '(t)= -1/2[1+16t^2]^(-3/2) [-2/3]32t
a ' (t)=[32t/3][1+16t^2]^(-3/2)

b '(t)=6J+K

Arthur
 
I made a error in the derivation of a'
a '(t) = -1/2 [1+16t^2]^9-3/2) 32t

the -2/3 multiplier in the original derivation was wrong
SORRY!!!

Arthur
 
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