The image shows an exercise from a textbook related to using the chain rule in partial differentiation to find the total derivative dtdz at a given point.
Problem Breakdown:
Part (a):
z=xcos(y),wherex=sin(t),y=cos(t)
Find dtdz.
Part (b):
z=x2+2xy+y2,wherex=etcos(t),y=etsin(t)
Find dtdz.
Solutions:
Part (a)
Expression for z:
z=xcos(y)
Given functions:
x=sin(t),y=cos(t)
The total derivative dtdz using the chain rule is given by:
dtdz=∂x∂zdtdx+∂y∂zdtdy
Partial derivatives:
∂x∂z=cos(y)∂y∂z=−xsin(y)
Derivatives of x and y with respect to t:
dtdx=cos(t)dtdy=−sin(t)
Substitute and simplify:
dtdz=cos(y)⋅cos(t)+(−xsin(y))⋅(−sin(t))dtdz=cos(cos(t))⋅cos(t)+sin(t)⋅sin(cos(t))⋅sin(t)dtdz=cos(cos(t))⋅cos(t)+sin2(t)⋅sin(cos(t))
Part (b)
Expression for z:
z=x2+2xy+y2
Given functions:
x=etcos(t),y=etsin(t)
Again, use the chain rule:
dtdz=∂x∂zdtdx+∂y∂zdtdy
Partial derivatives:
∂x∂z=2x+2y∂y∂z=2x+2y
Derivatives of x and y with respect to t:
dtdx=etcos(t)−etsin(t)dtdy=etsin(t)+etcos(t)
Substitute and simplify:
dtdz=(2x+2y)⋅dtdx+(2x+2y)⋅dtdy
Simplify further by substituting the values of x and y and their derivatives.
Would you like further steps for the simplification, or do you have any specific questions?
Follow-up Questions:
How does the chain rule apply when dealing with multiple variables?
Can you derive the partial derivatives of a more complex function?
What are the key differences between total and partial derivatives?
How would you handle a similar problem with more variables?
What happens if one of the functions is implicitly defined?
Tip: Always keep track of the variables being differentiated to avoid mistakes, especially when multiple variables depend on the same parameter.