Answer :
Answer:
se explanation, and the final value is 16.995574
Explanation:
a) As [tex]W[/tex] is a function of [tex]x(t), y(t), z(t)[/tex]. In order to use chain rule first we compute the partial derivatives of [tex]W[/tex] with respect to [tex]x(t), y(t), z(t)[/tex].
[tex]\frac{\partial W }{\partial x} = 7ye^{x}\\\frac{\partial W }{\partial y} = 7e^{x}\\\frac{\partial W }{\partial z} = -\frac{1}{z}[/tex]
then the total derivatives of each term with respect to t:
[tex]\frac{dx}{dt} =\frac{1}{t^+1}*2t=\frac{2t}{t^2 + 1}\\\frac{dy}{dt} = \frac{1}{t^2+1} \\\frac{dz}{dt} = e^t\\\\[/tex]
we get:
[tex]\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}\\\frac{dW}{dt} = 7ye^{x}\frac{2t}{t^2 + 1} + 7e^{x}\frac{1}{t^2+1}-\frac{1}{z}e^t[/tex]
next we replace x,y and z with the respective functions:
[tex]\frac{dW}{dt} = 7(tan^{-1}(t))e^{ln(t^2+1)}\frac{2t}{t^2 + 1} + 7e^{ln(t^2+1)}\frac{1}{t^2+1}-\frac{1}{e^t}e^t[/tex]
we cancel out the terms at the end, e^ln(expression) always will be equal to expression
after some algebra we get:
[tex]\frac{dW}{dt} = 7(tan^{-1}(t))e^{ln(t^2+1)}\frac{2t}{t^2 + 1} + 7e^{ln(t^2+1)}\frac{1}{t^2+1}-\frac{1}{e^t}e^t\\\frac{dW}{dt} = 14t(tan^{-1}(t))+6[/tex]
b) as a function of t we directly replace [tex]x(t), y(t), z(t)[/tex]. and derivate with respect to t, before derivating, we can cancel out some terms:
[tex]\frac{dw}{dt}= \frac{d}{dt} (7(tan^{-1}(t))*e^{ln(t^2+1)}-ln(e^{t}))\\\frac{dw}{dt}= \frac{d}{dt} (7(tan^{-1}(t))*(t^2+1)-t)= \frac{d}{dt}( 7t^2*tan^{-1}(t)+7tan^{-1}(t)-t)\\ \frac{dw}{dt}=14t*tan^{-1}(t) + 7t^2*\frac{1}{t^2+1}+ 7\frac{1}{t^2+1} -1\\\frac{dw}{dt}=14t*tan^{-1}(t) + 7(t^2+1)*\frac{1}{t^2+1} -1= 14t*tan^{-1}(t)+6[/tex]
now we replace t = 1
[tex]\frac{dw}{dt}(t=1)= 14*tan^{-1}(1)+6 = 6 +\frac{7\pi }{2} = 16.995574[/tex]