How to prove that F=(x^2-y^2)i-(2xy+y)j is a conservative force

Question : Prove that F=(x^2-y^2)i-(2xy+y)j is a conservative force Solution : To prove that Force\[\vec{F}=\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\] is a conservative force. We shall find curl of F \[ curl\left(\vec{F}\right)=\nabla\times\vec{F}=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}\right)\times\left(\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\right)\] \[=\left(\hat{i}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial…

how to Find curl of the vector F=(x^2-y^2)i-(2xy+y)j

Question : Find curl of the vector F=(x^2-y^2)i-(2xy+y)j Solution : Given vector \[\vec{F}=\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\] Then curl of F is \[ curl\left(\vec{F}\right)=\nabla\times\vec{F}=\left(\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}\right)\times\left(\left(x^2-y^2+x\right)\hat{i}-\left(2xy+y\right)\hat{j}\right)\] \[=\left(\hat{i}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial x}-\left(\hat{i}\times\hat{j}\right)\frac{\partial\left(2xy+y\right)}{\partial x}+\left(\hat{j}\times\hat{i}\right)\frac{\partial\left(x^2-y^2+x\right)}{\partial y}-\left(\hat{j}\times\hat{j}\right)\frac{\partial\left(2xy+y\right)}{\partial y}\] Since \[\hat{i}\times\hat{i}=\hat{j}\times\hat{j}=0\ ,\…

how to Find sinh(x+iy) formula

Question : Find sinh(x+iy) formula Solution : Let the complex number as \[z=sinh\left(x+iy\right)\] We know that \[\sin{\left(iz\right)}=isinh\left(z\right)\] So we shall multiply and divide by i we get \[z=\frac{isinh\left(x+iy\right)}{i}=\frac{\sin{\left(i\left(x+iy\right)\right)}}{i}=\frac{\sin{\left(ix-y\right)}}{i}\] Now we…

Solved Sum of the series 1/1.2-1/2.3+1/3.4-1/4.5+…

Question : Sum of the series 1/1.2-1/2.3+1/3.4-1/4.5+… Solution : Let the Sum of the series is \[S=\ \frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{3.4}-\frac{1}{4.5}+\ldots..\] \[=\left(\frac{1}{1}-\frac{1}{2}\right)-\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots.\] \[=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots.\right)+\left(-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots..\right)\] \[=\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots.\right)+\left(-1+1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\ldots..\right)\] We shall use formula of the series \[\log{\left(1+x\right)}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots..\] where…