Q. 56 - 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 $43–52.$

$f(x,y)=\frac{x}{y^2},P=(-4,7)$

Step-by-Step Solution

Answer

The answer for the equation is $3x+8y-3z=32$.

Step 1: Explanation

Given Information $f(x,y)=\frac{x}{y^2}$at position $P=\left(x_0,y_0\right)=(-4,7)$

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 $1$

$f_x(-4,7)(x+4)+f_y(-4,7)(y-7)=z-f(-4,7)$

Then,

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

$f_x(-4,7)={\left.\frac{1}{y^2}\right|}_{(-4,7)}$

Step 2: Substituting the equations

Equation $2$

$f_x(-4,7)=\frac{1}{49}$

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

$\begin{array}{r}{f}_y(-4,7)={\left.\frac{-2x}{y^3}\right|}_{(-4,7)}\\ \end{array}$

$f_y(-4,7)=\frac{-2\times (-4)}{7^3}$

Equation $3$

$f_y(-4,7)=\frac{8}{343}$

$f\left(x_0,y_0\right)=f(-4,7)=\frac{-4}{7^3}$

Equation $4$

$f(-4,7)=-\frac{4}{343}$

Step 3: Conclusion

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

$\begin{array}{r}\frac{1}{49}(x+4)+\frac{8}{343}(y-7)=z+\frac{4}{343}\end{array}$

$\begin{array}{r}\frac{1}{49}x+\frac{4}{49}+\frac{8}{343}y-\frac{56}{343}=z+\frac{4}{343}\\ \end{array}$

$\frac{1}{49}x+\frac{8}{343}y-z=\frac{4}{343}-\frac{4}{49}+\frac{56}{343}$

343 is multiply in both sides, we get,

$3x+8y-3z=4-28+56$

Finally we get ,

$3x+8y-3z=32$