Q. 59 - Calculus

Question

Use the first-order partial derivatives of the functions in Exercises $55–64$ to find the equation of the plane tangent to the graph of the function at the indicated point P. Note that these are the same functions as in Exercises $43–52.$

$f(x,y)=\sin \left(xy\right),P=\left(2,\frac{\pi }{2}\right)$

Step-by-Step Solution

Answer

The solution for the equation $\pi x+4y+2z=4\pi$.

Step 1: Introduction

Given Information$f(x,y)=\mathrm{Sin}\left(xy\right)$ at position $P=\left(x_0,y_0\right)=\left(2,\frac{\pi }{2}\right)$

The line of tangent equation is 

$f_x\left(x_0,y_0\right)\left(x-x_0\right)+f_y\left(x_0,y_0\right)\left(y-y_0\right)=z-f\left(x_0,y_0\right)$

Equation $1$

$f_x\left(2,\frac{\pi }{2}\right)(x-2)+f_y\left(2,\frac{\pi }{2}\right)\left(y-\frac{\pi }{2}\right)=z-f\left(2,\frac{\pi }{2}\right)$

Then,

$f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dx}\sin \left(xy\right)\right|}_{\left(x_0,y_0\right)}$

$f_x\left(x_0,y_0\right)={\left.\frac{\mathrm{\partial }}{\mathrm{\partial }x}(\sin (xy\left)\right)\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(xy\right)\right|}_{\left(x_0,y_0\right)}$

$\begin{array}{r}{f}_x\left(x_0,y_0\right)={\left.\cos \left(xy\right)\times y\right|}_{\left(x_0,y_0\right)}\\ f_x\left(x_0,y_0\right)={\left.y\times \cos \left(xy\right)\right|}_{\left(x_0,x_0\right)}\end{array}$

$f_x\left(2,\frac{\pi }{2}\right)=y\times \cos \left(xy\right)\left(2,\frac{\pi }{2}\right)$

$f_x\left(2,\frac{\pi }{2}\right)=\frac{\pi }{2}\cos \left(2\times \frac{\pi }{2}\right)$

$(\because \cos (\pi )=-1)$

$f_x\left(2,\frac{\pi }{2}\right)=-\frac{\pi }{2}$

Step 2: Equations 2 ,   3 and 4

Equation $2$

$f_x\left(2,\frac{\pi }{2}\right)=-\frac{\pi }{2}$

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dy}\sin \left(xy\right)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.\frac{\mathrm{\partial }}{\mathrm{\partial }y}(\sin (xy\left)\right)\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(xy\right)\right|}_{\left(x_0,y_0\right)}$

$\begin{array}{r}{f}_y\left(x_0,y_0\right)={\left.\cos \left(xy\right)\times x\right|}_{\left(x_0,y_6\right)}\\ \end{array}$

$f_y\left(x_0,y_0\right)={\left.x\times \cos \left(xy\right)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(2,\frac{\pi }{2}\right)=x\times \cos \left(xy\right)\mid \left(2,\frac{\pi }{2}\right)$

$f_y\left(2,\frac{\pi }{2}\right)=2\cos \left(2\times \frac{\pi }{2}\right)$

$f_y\left(2,\frac{\pi }{2}\right)=-2$

$(\because \cos (\pi )=-1)$

Equation $3$

$f_y\left(2,\frac{\pi }{2}\right)=-2$

$f\left(x_0,y_0\right)=f\left(2,\frac{\pi }{2}\right)=\sin \left(2\times \frac{\pi }{2}\right)$

$f\left(2,\frac{\pi }{2}\right)=\sin \left(\pi \right)$

Equation $4$

$f\left(2,\frac{\pi }{2}\right)=0$

Step 3: Conclusion

Equations $2, 3$ and $4$ are substituted by equation $1$, we get,

$-\frac{\pi }{2}(x-2)-2\left(y-\frac{\pi }{2}\right)=z-0$

$-\frac{\pi }{2}x+\frac{\pi }{2}\times 2-2y+2\times \frac{\pi }{2}=z$

$-\frac{\pi }{2}x-2y-z=-\pi -\pi$

$-\frac{\pi }{2}x-2y-z=-2\pi$

$2$ is multiplied by two sides,

$\pi x+4y+2z=4\pi$

Finally, the result for expression $\pi x+4y+2z=4\pi$