Q. 62 - Calculus

Question

$f(x,y)=\tan (x+y),P=(0,\pi )$

Step-by-Step Solution

Answer

 The Answer of the equation$f(x,y)=\tan (x+y),P=(0,\pi )$ is$x+y-z=\pi$

Step1:Given data

$f(x,y)=\tan (x+y)\text{ at point }P=\left(x_0,y_0\right)=(0,\pi )$

The line of tangent equation is

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

$f_x(0,\pi )(x-0)+f_y(0,\pi )(y-\pi )=z-f(0,\pi )$$\left(1\right)$

Step 2:Find fx

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

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

$f_x\left(x_0,y_0\right)={\left.{\sec }^2(x+y)\cdot 1\right|}_{\left(x_0,y_0\right)}$

$f_x\left(x_0,y_0\right)={\left.{\sec }^2(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_x(0,\pi )={\left.{\sec }^2(x+y)\right|}_{(0,\pi )}$

$f_x(0,\pi )={\sec }^2(0+\pi )$

$f_x(0,\pi )={\sec }^2\left(\pi \right)$

$f_x(0,\pi )=\frac{1}{{\cos }^2\left(\pi \right)}=\frac{1}{(\cos (\pi ){)}^2}$

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

$f_x(0,\pi )=\frac{1}{(-1{)}^2}$

$f_x(0,\pi )=1$$\left(2\right)$

Step 3: Find fy:

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

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

$f_y\left(x_0,y_0\right)={\left.{\sec }^2(x+y)\cdot 1\right|}_{\left(x_0{y}_0\right)}$

$f_y\left(x_0,y_0\right)={\left.{\sec }^2(x+y)\right|}_{\left(x_0,y_0\right)}$

$f_y(0,\pi )={\sec }^2(0+\pi )$

$f_y(0,\pi )={\sec }^2\left(\pi \right)$

$f_y(0,\pi )=\frac{1}{{\cos }^2\left(\pi \right)}=\frac{1}{(\cos (\pi ){)}^2}$

$(\because \cos (\pi )=-1)$$f_y(0,\pi )=\frac{1}{(-1{)}^2}$

$f_y(0,\pi )=1$$\left(3\right)$

Step4: Find f ( 0 , π )

$f\left(x_0,y_0\right)=f(0,\pi )=\tan (0+\pi )$

$f(0,\pi )=\tan \left(\pi \right)$

$f(0,\pi )=0$$\left(4\right)$

Step 5: f ( x , y ) = tan ⁡ ( x + y )

Substituting  equation $f_x(0,\pi )=1$,$f_y(0,\pi )=1$and $f(0,\pi )=0$in $f_x(0,\pi )(x-0)+f_y(0,\pi )(y-\pi )=z-f(0,\pi )$$1(x-0)+1(y-\pi )=z-0$get

$x-0+y-\pi =z$

$x+y-z=\pi$