Q. 60 - 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 $43–52.$

$f(x,y)=\cos \left(xy\right),P=(\pi ,-3)$

Step-by-Step Solution

Answer

The result for the equation is $z=-1$.

Step 1: Explanation

Already we have $f(x,y)=\cos \left(xy\right)$ at position $P=\left(x_0,y_0\right)=(\pi ,-3)$

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 $1$

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

Then,

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

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

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

$f_x(\pi ,-3)=-{\left.y\times \sin \left(xy\right)\right|}_{(\pi ,-3)}$

$f_x(\pi ,-3)=-(-3)\times \sin (\pi \times (-3\left)\right)$

$f_x(\pi ,-3)=3\times -\sin \left(3\pi \right)$

$f_x(\pi ,-3)=3\times 0$

$(\because \sin (3\pi )=0)$

Equation $2$

$f_x(\pi ,-3)=0$

$f_y\left(x_0,y_0\right)={\left.\frac{d}{dy}f(x,y)\right|}_{\left(x_0,3_0\right)}={\left.\frac{d}{dy}\cos \left(xy\right)\right|}_{\left(x_0,y_0\right)}$

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

$\begin{array}{r}{f}_y\left(x_0,y_0\right)=-{\left.\sin \left(xy\right)\times x\right|}_{\left(x_0,x_0\right)}\\ f_y\left(x_0,y_0\right)=-{\left.x\times \sin \left(xy\right)\right|}_{\left(x_0,y_5\right)}\end{array}$

$f_y(\pi ,-3)=-{\left.x\times \sin \left(xy\right)\right|}_{(\pi ,-3)}$

$f_y(\pi ,-3)=-\left(\pi \right)\times \sin (\pi \times (-3\left)\right)$

$f_y(\pi ,-3)=\pi \times -\sin \left(3\pi \right)$

$f_y(\pi ,-3)=\pi \times 0$

$(\because \sin (3\pi )=0)$

Equation $3$

$f_y(\mathrm{\pi },-3)=0$

$f\left(x_0,y_0\right)=f(\pi ,-3)=\cos (\pi \times (-3\left)\right)$

$f(\pi ,-3)=\cos (-3\pi )$

Equation $4$

$f(\mathrm{\pi },-3)=-1$

Step 2: Substitute in equations

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

We get

$0(x-\pi )+0(y+3)=z+1$

$0+0=z+1$

Step 3: Conclusion

Finally, the result is $z=-1$.