Math, asked by ayushbhandarkar10, 8 months ago

If w=xy+z with x=cos t,y=sin and z=t . Calculate dw/dt by using chain rule.

Answers

Answered by rashich1219
3

Given:

w = x y + z   where, x=cos t, y=sin t and z=t.

To Find:

Calculate dw/dt using chain rule.

Solution:

Since, w= x y + z  

Therefore,  

or , \[\begin{gathered}  dw/dt = \dfrac{d}{{dt}}(xy + z) \hfill \\   = d(xy)/dt + dz/dt \hfill \\   = y(dx/dt) + x(dy/dt) + d(t)/dt \hfill \\   = (\sin t)[d(\cos t)/dt] + (\cos t)[d(\sin t)/dt] + 1 \hfill \\   = \sin t( - \sin t) + \cos t(\cos t) + 1 \hfill \\   =  - {\sin ^2}t + {\cos ^2}t + 1 \hfill \\   = \cos 2t + 1 \hfill \\   = 1 + \cos 2t \hfill \\ \end{gathered} \]

Since, , \[ - {\sin ^2}t + {\cos ^2}t= cos 2t\]

Answered by pulakmath007
86

SOLUTION :

GIVEN

w = xy + z with x=cos t , y=sin t and z = t

TO DETERMINE

 \displaystyle \sf{ \frac{dw}{dt} \:  \: by \: chain \: rule  \: }

CONCEPT TO BE IMPLEMENTED

If w is a function of x, y, z and x, y, z are functions of t then By chain rule

 \displaystyle \sf{ \frac{dw}{dt} \:   =  \frac{ \partial w}{ \partial x}  \frac{dx}{dt} + \frac{ \partial w}{ \partial y}  \frac{dy}{dt}   + \frac{ \partial w}{ \partial z}  \frac{dz}{dt}\: }

EVALUATION

Here it is given that

w = xy + z

So

 \displaystyle \sf{  \frac{ \partial w}{ \partial x}  = y }

 \displaystyle \sf{  \frac{ \partial w}{ \partial y}  = x }

 \displaystyle \sf{  \frac{ \partial w}{ \partial z}  = 1}

Also

x=cos t , y=sin t and z = t

So

 \displaystyle \sf{ \frac{dx}{dt} \:   =  -  \sin t\: }

 \displaystyle \sf{ \frac{dy}{dt} \:   =   \cos t\: }

 \displaystyle \sf{ \frac{dz}{dt} \:   = 1 }

Hence by the Chain Rule

 \displaystyle \sf{ \frac{dw}{dt} \:   =  \frac{ \partial w}{ \partial x}  \frac{dx}{dt} + \frac{ \partial w}{ \partial y}  \frac{dy}{dt}   + \frac{ \partial w}{ \partial z}  \frac{dz}{dt}\: }

\implies \displaystyle \sf{ \frac{dw}{dt} =  - y \sin t + x  \cos t + 1 }

\implies \displaystyle \sf{ \frac{dw}{dt} = 1  +  { \cos}^{2} t -  { \sin}^{2}t  }

\implies \displaystyle \sf{ \frac{dw}{dt} = 1  +  { \cos}2 t }

\implies \displaystyle \sf{ \frac{dw}{dt} = 2 { \cos}^{2} t   }

\implies \displaystyle \sf{ \frac{dw}{dt} = 2 {x}^{2}   }

FINAL RESULT

\boxed{  \:  \: \displaystyle \sf{ \frac{dw}{dt} = 2 {x}^{2} \:  \:    }}

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