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Question: The potential energy of a force field \[\vec F\] is given by \(U(x,y) = \sin (x + y)\). The force ac...

The potential energy of a force field F⃗\vec F is given by U(x,y)=sin⁡(x+y)U(x,y) = \sin (x + y). The force acting on the particle of mass m at (0,π4)\left( {0,\dfrac{\pi }{4}} \right) is
A) 1
B) 2\sqrt 2
C) 12\dfrac{1}{{\sqrt 2 }}
D) 0

Explanation

Solution

Here we have to use the concept of the partial derivative. The partial derivative is a derivative of a function of two or more variables with respect to one variable, the other(s) will be treated as constant. Partial derivatives are used in vector calculus and differential geometry. Taking the partial derivatives of the given equation and then using the point given will give us a very easy solution to the question in hand. For this question, we have to use the partial derivative equation.

Formula Used: For example, if we have an equation f(x,y)=x2yf(x,y) = {x^2}y.
Then taking partial derivatives we get,
∂f∂x=∂∂xx2y\dfrac{{\partial f}}{{\partial x}} = \dfrac{\partial }{{\partial x}}{x^2}y, if we consider y as constant, we get,
∂f∂x=2xy\dfrac{{\partial f}}{{\partial x}} = 2xy.
Again, from the equation
∂f∂x=∂∂xx2y\dfrac{{\partial f}}{{\partial x}} = \dfrac{\partial }{{\partial x}}{x^2}y, if we consider x as constant, we get,
∂f∂x=x2.1\dfrac{{\partial f}}{{\partial x}} = {x^2}.1

Complete step by step answer:
Given in the equation we have, Force F⃗\vec F is
U(x,y)=sin⁡(x+y)U(x,y) =\sin (x + y).
Here, taking partial derivatives we get,
F⃗=−∂U∂xi^−−∂U∂xj^\vec F = \dfrac{{ - \partial U}}{{\partial x}}\hat i - \dfrac{{ - \partial U}}{{\partial x}}\hat j.
This will further give us F⃗=−cos⁡(x+y)i^−cos⁡(x+y)j^\vec F = - \cos (x + y)\hat i - \cos (x + y)\hat j.
Further equating we get at the point (0,π4)\left( {0,\dfrac{\pi }{4}} \right),
F⃗=−cos⁡(π4)i^−cos⁡(π4)j^\vec F = - \cos (\dfrac{\pi }{4})\hat i - \cos (\dfrac{\pi }{4})\hat j .
Putting the values of cos⁡(π4)\cos (\dfrac{\pi }{4}), we have,
F⃗=−12i^−12j^=−12(i^+j^)\vec F = - \dfrac{1}{{\sqrt 2 }}\hat i - \dfrac{1}{{\sqrt 2 }}\hat j = - \dfrac{1}{{\sqrt 2 }}(\hat i + \hat j).
Now the value of i^+j^=12+12=2\hat i + \hat j = \sqrt {{1^2} + {1^2}} = \sqrt 2 as i^=(−1)2\hat i = {( - 1)^2} , and j^=(−1)2\hat j = {( - 1)^2}.
This will finally give us,
∣F⃗∣=∣−12(i^+j^)∣=∣−12∣×2=1\left| {\vec F} \right| = \left| { - \dfrac{1}{{\sqrt 2 }}(\hat i + \hat j)} \right| = \left| { - \dfrac{1}{{\sqrt 2 }}} \right| \times \sqrt 2 = 1 as ∣−12∣=12\left| { - \dfrac{1}{{\sqrt 2 }}} \right| = \dfrac{1}{{\sqrt 2 }}.
Thus, our final answer comes out to be 1N which is option number (A).

Note: Alternate method to solve the question. This method is a concise or shorter method to solve the same question. If we are solving a numerical using this method is beneficial but when giving a subjective answer, we should follow the first method.
Given the force F⃗\vec Fas
U(x,y)=sin⁡(x+y)U(x,y) =\sin (x + y).
Taking derivative, we get,
F⃗=−dUdx−dsin⁡(x+y)dx=cos⁡(x+y)\vec F = - \dfrac{{dU}}{{dx}} - \dfrac{{d\sin (x + y)}}{{dx}} = \cos (x + y).
For the given point (0,π4)\left( {0,\dfrac{\pi }{4}} \right) can rewrite the equation,
F⃗=cos⁡(x+y)=cos⁡(0+π4)=cos⁡(π4)=1\vec F = \cos (x + y) = \cos (0 + \dfrac{\pi }{4}) = \cos (\dfrac{\pi }{4}) = 1.
So, the final result we get is 1 which is the option (A).