Question

For each of the following vector fields F , decide whether it is conservative or not by computing the appropriate first order partial derivatives. Type in a potential function f (that is, ∇f=F) with f(0,0)=0. If it is not conservative, type N.

A. F(x,y) = (-16x+2y)i + (2x+10y)j
f(x,y) =
B. F(x,y)= -8yi - 7xj
f(x,y) =
C. F(x,y) = (-8 sin y)i + (4y - 8x cos y)j
f(x,y) =

Answer 1

Answer:

• A. F is conservative, and $f(x,y)=-8x^2 +2xy+5y^2$
• B. N
• C. F is conservative, and $f(x,y)=-8x\sin y +2y^2$

Step-by-step explanation:

Remember that a vector field

$F(x,y)=G(x,y)\uvec{i} + H(x,y)\uvec{j}$

is conservative if $\frac{\partial G}{\partial y} = \frac{\partial H}{\partial x}$.

• A. In this case, $G(x,y)= -16x+2y$ and  $H(x,y)= 2x+10y$. The partial derivatives are $\frac{\partial G}{\partial y}=2$ and  $\frac{\partial H}{\partial x}=2$. Note that $\frac{\partial G}{\partial y} = \frac{\partial H}{\partial x}$ therefore F is conservative. By definition, a potential function f must satisfy $\frac{\partial f}{\partial x}=G(x,y)=-16x+2y$ and $\frac{\partial f}{\partial y}=H(x,y)=2x+10y$. Integrating respect to x on the first equation, $f(x,y)=\int -16x+2y dx =-8x^2 +2xy+ c(y)$ where c(y) is a function constant on x. To find c(y), we differentiate respect to y and use the previous equation to obtain $\frac{\partial f}{\partial y}=2x+ c'(y)=2x+10y$. Then $c'(y)= 10y$ so $c(y)= 5y^2+k$ for some constant k. Now $f(x,y)=-8x^2 +2xy+5y^2+k$. We need that f(0,0)=0 so $0=f(0,0)=-8(0)+2(0)(0)+5(0)+k$ then k=0 and $f(x,y)=-8x^+2xy+5y^2$  

• B. Here,  $G(x,y)=-8y$ and  $H(x,y)= -7x$. Because the partial derivatives $\frac{\partial G}{\partial y}=-8$ and  $\frac{\partial H}{\partial x}=-7$ are not equal, F is not conservative.

• C. We have that $G(x,y)= -8\sin y$ and  $H(x,y)= 4y-8x\cos y$, so the partial derivatives are $\frac{\partial G}{\partial y}=-8\cos y$ and  $\frac{\partial H}{\partial x}=-8cos y$. The partial derivartives are equal, therefore F is conservative. Using the same method from part A, a potential function f satisfies $\frac{\partial f}{\partial x}=-8\sin y$ and $\frac{\partial f}{\partial y}= 4y-8x\cos y$. Integrate respect to x on the first equation to get $f(x,y)=\int -8\sin y dx =-8x\sin y+ b(y)$ for a function b(y) constant in x. Now differentiate respect to y to obtain $\frac{\partial f}{\partial y}=-8x\cos y+ b'(y)=4y-8x\cos y$. Then $b'(y)= 4y$ so $b(y)= 2y^2+t$ for a constant t. Because f(0,0)=0, $0=-8(0)\sin(0)+2(0)+t$, then t=0 and $f(x,y)=-8x\sin y +2y^2$.

Answer 2

It can be noted that F is conservative from the derivative computed and f(x,y) = 8x² + 2xy - 5y².

How to calculate the derivative?

From the information given, G(x, y) = -16x + 2y and H(x,y) = 2x + 10y.

Then, it is important to calculate the partial derivative which will be 2. Therefore, F is conservative. Therefore, the integration of the value will be:

= -16x + 2ydx

= -8x² + 2xy + c(y

where c(y) = 5y²

In conclusion, the correct option is f(x,y) = 8x² + 2xy - 5y².

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