Q. 57 - 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 $43–52.$

$f(x,y)=\frac{x}{x^2+y^2},P=(-3,0)$

Step-by-Step Solution

Answer

The answer for the equation $x+9z=-6$

Step 1: Explanation

Given Information$f(x,y)=\frac{x}{x^2+y^2}$ at position $P=\left(x_0,y_0\right)=(-3,0)$

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 $1$

$f_x(-3,0)(x+3)+f_y(-3,0)(y-0)=z-f(-3,0)$

Then,

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

$f_x\left(x_0,y_0\right)=\frac{\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(x\right)\left(x^2+y^2\right)-\frac{\mathrm{\partial }}{\mathrm{\partial }x}\left(x^2+y^2\right)x}{{\left(x^2+y^2\right)}^2}$

$f_x\left(x_0,y_0\right)={\left.\frac{1\times \left(x^2+y^2\right)-2xx}{{\left(x^2+y^2\right)}^2}\right|}_{\left(x_0,y_0\right)}$

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

$f_x(-3,0)={\left.\frac{-x^2+y^2}{{\left(x^2+y^2\right)}^2}\right|}_{(-3,0)}$

$f_x(-3,0)=\frac{-(-3{)}^2+0^2}{{\left((-3{)}^2+0^2\right)}^2}$

$f_x(-3,0)=\frac{-9}{9^2}$

Step 2: Equations 2 ,   3 and 4

Equation $2$

$f_x(-3,0)=\frac{-1}{9}$

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

$\begin{array}{r}{f}_y\left(x_0,y_0\right)={\left.x\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(\frac{1}{x^2+y^2}\right)\right|}_{\left(x_0,y_0\right)}\\ \end{array}$

$f_y\left(x_0,y_0\right)={\left.x\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left({\left(x^2+y^2\right)}^{-1}\right)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.x\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left({\left(x^2+y^2\right)}^{-1}\right)\frac{\mathrm{\partial }}{\mathrm{\partial }y}\left(x^2+y^2\right)\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)={\left.x\left(-\frac{1}{{\left(x^2+y^2\right)}^2}\right)\times 2y\right|}_{\left(x_0,y_0\right)}$

$f_y\left(x_0,y_0\right)=-{\left.\frac{2xy}{{\left(x^2+y^2\right)}^2}\right|}_{\left(x_0{y}_0\right)}$

$f_y(-3,0)=\frac{-2(-3)0}{{\left((-3{)}^2+0^2\right)}^2}$

Equation $3$

$f_y(-3,0)=0$

$f\left(x_0,y_0\right)=f(-3,0)=\frac{-3}{(-3{)}^2+0^2}$

Equation $4$

$f(-3,0)=-\frac{1}{3}$

Step 3: Substitute in Equations

Substitute equation $2,3$ and $4$ in equation $1$, we get

$-\frac{1}{9}(x+3)+0(y-0)=z+\frac{1}{3}$

$-\frac{1}{9}x-\frac{3}{9}+0=z+\frac{1}{3}$

$-\frac{1}{9}x-z=\frac{1}{3}+\frac{3}{9}$

$9$ is multiplied by two sides, we get

$-x-9z=3+3$