PLEASEE SHOW WORK ON ALL (except #3 but also answer :D) 10 POINTS EACH QUESTION SO 50 POINTS + BRAINLY

Determine whether or not F is a conservative vector field. If it is, find a function f such that F = ∇f. (If the vector field is

kodGreya [7K]

We're looking for a scalar function $f(x,y)$ such that $\nabla f(x,y)=\vec F(x,y)$ . That is,

$\dfrac{\partial f}{\partial x}=2x-6y$

$\dfrac{\partial f}{\partial y}=-6x+6y-7$

Integrate the first equation with respect to $x$ :

$f(x,y)=x^2-6xy+g(y)$

Differentiate with respect to $y$ :

$-6x+6y-7=-6x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=6y-7$

Integrate with respect to $y$ :

$g(y)=3y^2-7y+C$

So $\vec F$ is indeed conservative with the scalar potential function

$f(x,y)=x^2-6xy+3y^2-7y+C$

where $C$ is an arbitrary constant.

6 0

3 years ago

Which of the following describes the line with equation y=-5

slava [35]

Answer: please post your answer choices and then I can help

Step-by-step explanation:

8 0

3 years ago

How to find the derivative of cos^2x? i seem to be confused.

slamgirl [31]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2927231

————————

You can actually use either the product rule or the chain rule for this one. Observe:

• Method I:

y = cos² x

y = cos x · cos x

Differentiate it by applying the product rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x\cdot cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x)\cdot cos\,x+cos\,x\cdot \dfrac{d}{dx}(cos\,x)}$

The derivative of cos x is – sin x. So you have

$\mathsf{\dfrac{dy}{dx}=(-sin\,x)\cdot cos\,x+cos\,x\cdot (-sin\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=-sin\,x\cdot cos\,x-cos\,x\cdot sin\,x}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

—————

• Method II:

You can also treat y as a composite function:

$\left\{\!
\begin{array}{l}
\mathsf{y=u^2}\\\\
\mathsf{u=cos\,x}
\end{array}
\right.$

and then, differentiate y by applying the chain rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\
\mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(u^2)\cdot \dfrac{d}{dx}(cos\,x)}$

For that first derivative with respect to u, just use the power rule, then you have

$\mathsf{\dfrac{dy}{dx}=2u^{2-1}\cdot \dfrac{d}{dx}(cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=2u\cdot (-sin\,x)\qquad\quad (but~~u=cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=2\,cos\,x\cdot (-sin\,x)}$

and then you get the same answer:

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: <em>derivative chain rule product rule composite function trigonometric trig squared cosine cos differential integral calculus</em>

3 0

3 years ago

How can i find the surface area of a rectangle

omeli [17]

Answer:

Legenth times width

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

What is the slope of (-5,3) and (7,9)

Softa [21]

Answer:

-35

___

27

Step-by-step explanation:

Start by setting up your pairs (x,y)

-5 & 7 are x

3 & 9 are y

X * X

---------- ÷

Y * Y

(-5, 7)

---------

(3, 9)

-35

-------

27

3 0

2 years ago