Kushwoho44 said:
Homework Statement
Let x=ts^2 -1 and y=ln(s)-t
Use the chain rule for functions of two variables to determine ∂f/∂t at (s,t)=(1,1)
What is f? Are we to assume that f is some differentiable function of x and y?
The Attempt at a Solution
y=ln(s)-t
∂f/∂t= ∂f/∂s X ∂s/∂t -1
\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}
\frac{\partial f}{\partial s}= \frac{\partial f}{\partial x}(2st)+ \frac{\partial f}{\partial y}\frac{1}{s}
At (s, t)= (1, 1), that would be
\frac{\partial f}{\partial s}(1, 1)= 2\frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}
I have no clue where this is coming from. Before, x and y were functions of s and t. Now t is a function of x and s? Is this a different problem?
∂t/∂s= -2(x+1)/s^3
∂s/∂t=s^3/-2(x+1)
∴ ∂f/∂t= s^2/-2(x+1) -1
= ∂f/∂t= s^2/-2(ts^2) -1
∂f/∂t (1,1) = 1/-2 +1
=0.5
However, the answer is 1 and I have spent two hours trying to work out why I am wrong. All help will be met with great appreciation. Thanks <3
Given f a function of x and y and x=ts^2 -1 and y=ln(s)-t as before
\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}= s^2\frac{\partial f}{\partial x}- \frac{\partial f}{\partial y}
At (s, t)= (1, 1) that is
\frac{\partial f}{\partial t}= \frac{\partial f}{\partial x}- \frac{\partial f}{\partial y}
It's
impossible to give specific numbers, as you seem to have, because you have not given f as a function of x and y.
