Annabel Zamora

2023-04-01

An object moving in the xy-plane is acted on by a conservative force described by the potential energy function $U(x,y)=\alpha (\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}})$ where $\alpha$ is a positive constant. Derivative an expression for the force expressed terms of the unit vectors $\overrightarrow{i}$ and $\overrightarrow{j}$ .

Jeffery Romero

Beginner2023-04-02Added 5 answers

To find the force $\overrightarrow{F}$ associated with the given potential-energy function $U(x,y)=a(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}})$, we can use the relationship between force and potential energy. The force $\overrightarrow{F}$ can be expressed as the negative gradient of the potential energy function:

$\overrightarrow{F}=-\nabla U(x,y)$

Taking the partial derivatives of $U(x,y)$ with respect to $x$ and $y$, we obtain:

$\frac{\partial U}{\partial x}=\frac{\partial}{\partial x}\left(a(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}})\right)=-\frac{2a}{{x}^{3}}$

$\frac{\partial U}{\partial y}=\frac{\partial}{\partial y}\left(a(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}})\right)=-\frac{2a}{{y}^{3}}$

The negative sign indicates that the force acts in the opposite direction of the gradient.

Now, we can express the force $\overrightarrow{F}$ in terms of the unit vectors $\hat{i}$ and $\hat{j}$:

$\overrightarrow{F}=\frac{2a}{{x}^{3}}\hat{i}+\frac{2a}{{y}^{3}}\hat{j}$

Therefore, the expression for the force $\overrightarrow{F}$ in terms of the unit vectors $\hat{i}$ and $\hat{j}$ is:

$\overrightarrow{F}=\frac{2a}{{x}^{3}}\hat{i}+\frac{2a}{{y}^{3}}\hat{j}$

$\overrightarrow{F}=-\nabla U(x,y)$

Taking the partial derivatives of $U(x,y)$ with respect to $x$ and $y$, we obtain:

$\frac{\partial U}{\partial x}=\frac{\partial}{\partial x}\left(a(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}})\right)=-\frac{2a}{{x}^{3}}$

$\frac{\partial U}{\partial y}=\frac{\partial}{\partial y}\left(a(\frac{1}{{x}^{2}}+\frac{1}{{y}^{2}})\right)=-\frac{2a}{{y}^{3}}$

The negative sign indicates that the force acts in the opposite direction of the gradient.

Now, we can express the force $\overrightarrow{F}$ in terms of the unit vectors $\hat{i}$ and $\hat{j}$:

$\overrightarrow{F}=\frac{2a}{{x}^{3}}\hat{i}+\frac{2a}{{y}^{3}}\hat{j}$

Therefore, the expression for the force $\overrightarrow{F}$ in terms of the unit vectors $\hat{i}$ and $\hat{j}$ is:

$\overrightarrow{F}=\frac{2a}{{x}^{3}}\hat{i}+\frac{2a}{{y}^{3}}\hat{j}$

I need to find a unique description of Nul A, namely by listing the vectors that measure the null space

?

$A=\left[\begin{array}{ccccc}1& 5& -4& -3& 1\\ 0& 1& -2& 1& 0\\ 0& 0& 0& 0& 0\end{array}\right]$T must be a linear transformation, we assume. Can u find the T standard matrix.$T:{\mathbb{R}}^{2}\to {\mathbb{R}}^{4},T\left({e}_{1}\right)=(3,1,3,1)\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}T\left({e}_{2}\right)=(-5,2,0,0),\text{}where\text{}{e}_{1}=(1,0)\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{e}_{2}=(0,1)$

?Find a nonzero vector orthogonal to the plane through the points P, Q, and R. and area of the triangle PQR

Consider the points below

P(1,0,1) , Q(-2,1,4) , R(7,2,7).

a) Find a nonzero vector orthogonal to the plane through the points P,Q and R.

b) Find the area of the triangle PQR.Consider two vectors A=3i - 1j and B = - i - 5j, how do you calculate A - B?

Let vectors A=(1,0,-3) ,B=(-2,5,1) and C=(3,1,1), how do you calculate 2A-3(B-C)?

What is the projection of $<6,5,3>$ onto $<2,-1,8>$?

What is the dot product of $<1,-4,5>$ and $<-5,7,3>$?

Which of the following is not a vector quantity?

A)Weight;

B)Nuclear spin;

C)Momentum;

D)Potential energyHow to find all unit vectors normal to the plane which contains the points $(0,1,1),(1,-1,0)$, and $(1,0,2)$?

What is a rank $1$ matrix?

How to find unit vector perpendicular to plane: 6x-2y+3z+8=0?

Can we say that a zero matrix is invertible?

How do I find the sum of three vectors?

How do I find the vertical component of a vector?

How would you find a unit vector perpendicular to plane ABC where points are A(3,-1,2), B(1,-1,-3) and C(4,-3,1)?