calculus 3

Question

express dw/dt as a function of t, both by using the chain rule and by expressing w in terms of t and differentiating directly with respect to t. then, evaluate dw/dt at the given value of t.

 

1. w=x/z+y/z: x=cos²t, y=sin²t, z=1/t, t=3

 

2. w=2ye^x-ln z; x=ln(t^2+1), tan^-1t, z=e^t, t=1

Answer 1

If you substitute first you get w=tcos²(t)+tsin²(t)=t. Therefore w'=1.

The other way you have to use the chain rule:

dw/dt= dw/dx*dx/dt+dw/dy*dy/dt+dw/dz*dz/dt=

=1/z*(-2cost*sint)+ 1/z*(2sint cost)+(x+y)*((-1)/z²)*(-1/t²)=(x+y)*(t²/t²)=x+y=1.

 

Note: we use * for multiplication here.

For more info on partial derivatives check out:

 

http://en.wikipedia.org/wiki/Chain_rule

 

I hope this helps.

Answer 2

1-----

 

w=(x+y)/z

   =[cos(t)^2+ sin(t)^2]/(1/t)

 

   =t

dw/dt=1

 

2---------

 

dw/dt=2y(e^x)2t/(t^2+1) +2(e^x)*(1/(1+t^2) -[1/(e^t)]*(e^t)

 

dw/dt= 2taninverse(t)*2t+2-1          ,   t=1 given

 

dw/dt= 4t* taninverse(t)+2-1

 

dw/dt=4*pi/4+1

 

         =pi+1