Pick any exercise in Chapter 12 of the calculus textbook.
Post the solution or your attempt.
Discuss any issues you might have had.
In Exercises 27 – 30, form a function \(z = f(x;y)\) such that \(f_x\) and \(f_y\) match those given.
For a solution to exist, we must have \(f_{xy}=f_{yx}\) .
Checking this: \(f_{xy} = \cos(y) =f_{yx}\) .
OK, we can proceed.
\(f_x = \sin (y) + 1\), so
\[\begin{aligned} z = f(x;y) &= \int{[f_x] dx} \\ &= \int[\sin (y) + 1]dx \\ &= x\sin (y) + x + h(y)+ C \end{aligned}\]
where \(h\) is any function of \(y\) (but not \(x\)).
\(f_y = x \cos (y)\)
\[\begin{aligned} z = f(x;y) &= \int [f_y] dy \\ &= \int [x \cos (y)] dy \\ &= x\sin (y) + g(x) + C \end{aligned}\]
where \(g\) is any function of \(x\) (but not \(y\)).
So, we need to have:
\[x\sin (y) + x + h(y)+ C = x\sin (y) + g(x) + C\]
which works when \(g(x)=x\) and \(h(y)=0\) .
Therefore, the solution is \(z = f(x;y) = \boxed{x\sin (y) + x + C}\) .
For a solution to exist, we must have \(f_{xy}=f_{yx}\) .
Checking this: \(f_{xy} = 1 =f_{yx}\) .
OK, we can proceed.
\(f_x = x + y\), so
\[\begin{aligned} z = f(x;y) &= \int{[f_x] dx} \\ &= \int[ x + y]dx \\ &= \frac{x^2}{2} + xy + h(y)+ C \end{aligned}\]
where \(h\) is any function of \(y\) (but not \(x\)).
\(f_y = x \cos (y)\)
\[\begin{aligned} z = f(x;y) &= \int [f_y] dy \\ &= \int[ x + y]dy \\ &= xy + \frac{y^2}{2} + g(x)+ C \end{aligned}\]
where \(g\) is any function of \(x\) (but not \(y\)).
So, we need to have:
\[\frac{x^2}{2} + xy + h(y)+ C = xy + \frac{y^2}{2} + g(x)+ C\]
which works when \(g(x)=\frac{x^2}{2}\) and \(h(y)=\frac{y^2}{2}\) .
Therefore, the solution is \[\begin{aligned} z=f(x;y) &=\frac{x^2}{2} + xy + \frac{y^2}{2} + C \\ &= \boxed{\frac{1}{2} (x+y)^2 + C} \end{aligned}\] .
For a solution to exist, we must have \(f_{xy}=f_{yx}\) .
Checking this: \(f_{xy} = 6x-8y =f_{yx}\) .
OK, we can proceed.
\(f_x = 6xy - 4y^2\), so
\[\begin{aligned} z = f(x;y) &= \int{[f_x] dx} \\ &= \int[6xy - 4y^2]dx \\ &= 3x^2y -4xy^2 + h(y)+ C \end{aligned}\]
where \(h\) is any function of \(y\) (but not \(x\)).
\(f_y = 3x^2 - 8xy + 2\)
\[\begin{aligned} z = f(x;y) &= \int [f_y] dy \\ &= \int[ 3x^2 - 8xy + 2]dy \\ &= 3x^2y -4xy^2 +2y + g(x)+ C \end{aligned}\]
where \(g\) is any function of \(x\) (but not \(y\)).
So, we need to have:
\[3x^2y -4xy^2 + h(y)+ C = 3x^2y -4xy^2 +2y + g(x)+ C\]
which works when \(g(x)=0\) and \(h(y)=2y\) .
The solution is \(z=f(x; y) = \boxed{3x^2y - 4xy^2 + 2y + C}\) .
For a solution to exist, we must have \(f_{xy}=f_{yx}\) .
Checking this: \(f_{xy} = \frac{-2x2y}{(x^2+y^2)^2} =f_{yx}\) .
OK, we can proceed.
\(f_x = \frac{2x}{x^2 + y^2}\), so
\[\begin{aligned} z = f(x;y) &= \int{[f_x] dx} \\ &= \int\left[ \frac{2x}{x^2 + y^2}\right]dx \\ &= \ln\left(x^2 + y^2\right) + h(y)+ C \end{aligned}\]
where \(h\) is any function of \(y\) (but not \(x\)).
\(f_y = x \cos (y)\)
\[\begin{aligned} z = f(x;y) &= \int [f_y] dy \\ &= \int\left[ \frac{2y}{x^2 + y^2}\right]dy \\ &= \ln\left(x^2 + y^2\right) + g(x)+ C \end{aligned}\]
where \(g\) is any function of \(x\) (but not \(y\)).
So, we need to have:
\[\ln\left(x^2 + y^2\right)+ h(y)+ C = \ln\left(x^2 + y^2\right) + g(x)+ C\]
which works when \(g(x)=0=h(y)\) .
Therefore, the solution is \(z=f(x;y)=\boxed{\ln\left(x^2 + y^2\right) + C}\) .
It was a very useful review of material which I had first learned a very long time ago.