Q. 58 - 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 $43–52.$

$f(x,y)=\frac{x}{x^2+y^2-1},P=(1,-3)$

Step-by-Step Solution

Answer

The answer for the equation $7x+6y-9z=-12$.

Step 1: Explanation

Given Information, $f(x,y)=\frac{x}{x^2+y^2-1}$at position $P=\left(x_0,y_0\right)=(1,-3)$

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 $1$

$f_x(1,-3)(x-1)+f_y(1,-3)(y+3)=z-f(1,-3)$

Then,

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

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

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

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

$f_x(1,-3)={\left.\frac{-x^2+y^2-1}{{\left(x^2+y^2-1\right)}^2}\right|}_{(1,-3)}$

$f_x(1,-3)=\frac{-(1{)}^2+(-3{)}^2-1}{{\left((1{)}^2+(-3{)}^2-1\right)}^2}$

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

Step 2: Equations 2 ,   3 and 4

Equation $2$

$f_x(1,-3)=\frac{7}{9}$

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

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

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

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

$f_y\left(x_0,y_0\right)={\left.x\left(-\frac{1}{{\left(x^2+y^2-1\right)}^2}\right)\times 2y\right|}_{\left(x_0,5_0\right)}$

$f_y(1,-3)=-{\left.\frac{2xy}{{\left(x^2+y^2-1\right)}^2}\right|}_{(1,-3)}$

$f_y(1,-3)=-\frac{2\times 1\times (-3)}{{\left((1{)}^2+(-3{)}^2-1\right)}^2}$

$f_y(1,-3)=-\frac{-6}{1+9-1}$

Equation $3$

$f_y(1,-3)=\frac{6}{9}$

$f\left(x_0,y_0\right)=f(1,-3)=\frac{1}{1^2+(-3{)}^2-1}$

Equation $4$

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

Step 3: Conclusion

Equation $2,3$ and $4$ are substituted in equation $1$, we get,

$\frac{7}{9}(x-1)+\frac{6}{9}(y+3)=z-\frac{1}{9}$

$\frac{7}{9}x-\frac{7}{9}+\frac{6}{9}y+\frac{18}{9}=z-\frac{1}{9}$

$\frac{7}{9}x+\frac{6}{9}y-z=-\frac{1}{9}+\frac{7}{9}-\frac{18}{9}$

$\frac{7}{9}x+\frac{6}{9}y-z=-\frac{12}{9}$

$9$ is multiplied by two sides, we get,

$7x+6y-9z=-12$

Finally, the result is $7x+6y-9z=-12$.