Q. 61 - 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 $43–52.$

$f(x,y)=\tan \left(xy\right),P=\left(1,-\frac{\pi }{4}\right)$

Step-by-Step Solution

Answer

The answer for the equation $\pi x-4y+2z=2\pi -2$.

Step 1: Explanation

Given Information, $f(x,y)=\tan \left(xy\right)$ at position $P=\left(x_0,y_0\right)=\left(1,-\frac{\pi }{4}\right)$

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 $1$

$f_x\left(1,-\frac{\pi }{4}\right)(x-1)+f_y\left(1,-\frac{\pi }{4}\right)\left(y+\frac{\pi }{4}\right)=z-f\left(1,-\frac{\pi }{4}\right)$

Then,

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

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

$\begin{array}{r}{f}_x\left(x_0,y_0\right)={\left.{\sec }^2\left(xy\right)\times y\right|}_{\left(x_0,y_0\right)}\\ f_x\left(x_0,y_0\right)={\left.y\times {\sec }^2\left(xy\right)\right|}_{\left(x_0,y_0\right)}\end{array}$

$f_x\left(1,-\frac{\pi }{4}\right)=y\times {\sec }^2\left(xy\right)\mid \left(1,-\frac{\pi }{4}\right)$

$f_x\left(1,-\frac{\pi }{4}\right)=-\frac{\pi }{4}\times {\sec }^2\left(1\times \left(-\frac{\pi }{4}\right)\right)$

$f_x\left(1,-\frac{\pi }{4}\right)=-\frac{\pi }{4}\times {\sec }^2\left(-\frac{\pi }{4}\right)$

$f_x\left(1,-\frac{\pi }{4}\right)=-\frac{\pi }{4}\left(\frac{1}{{\cos }^2\left(-\frac{\pi }{4}\right)}\right)=-\frac{\pi }{4}\left(\frac{1}{{\left(\cos \left(\frac{\pi }{4}\right)\right)}^2}\right)$

$f_x\left(1,-\frac{\pi }{4}\right)=-\frac{\pi }{4}\left(\frac{1}{{\left(\frac{\sqrt{2}}{2}\right)}^2}\right)$

$\left(\because \cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\right)$

$f_x\left(1,-\frac{\pi }{4}\right)=-\frac{\pi }{4{\left(\frac{\sqrt{2}}{2}\right)}^2}=-\frac{\pi }{4\frac{1}{2}}$

Step 2: Equations 2 ,   3 and 4

Equation $2$

$f_x\left(1,-\frac{\pi }{4}\right)=-\frac{\pi }{2}$

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

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

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

$f_y\left(1,-\frac{\pi }{4}\right)=1\times {\sec }^2\left(1\times \left(-\frac{\pi }{4}\right)\right)$

$f_y\left(1,-\frac{\pi }{4}\right)={\sec }^2\left(-\frac{\pi }{4}\right)$

$f_y\left(1,-\frac{\pi }{4}\right)=\left(\frac{1}{{\cos }^2\left(-\frac{\pi }{4}\right)}\right)=\left(\frac{1}{{\left(\cos \left(\frac{\pi }{4}\right)\right)}^2}\right)$

$f_y\left(1,-\frac{\pi }{4}\right)=\left(\frac{1}{{\left(\frac{\sqrt{2}}{2}\right)}^2}\right)$

$\left(\because \cos \left(\frac{\pi }{4}\right)=\frac{\sqrt{2}}{2}\right)$

$f_y\left(1,-\frac{\pi }{4}\right)=\frac{1}{{\left(\frac{\sqrt{2}}{2}\right)}^2}=\frac{1}{\frac{1}{2}}$

Equation $3$

$f_y\left(1,-\frac{\pi }{4}\right)=2$

$f\left(x_0,y_0\right)=f\left(1,-\frac{\pi }{4}\right)=\tan \left(1\times \left(-\frac{\pi }{4}\right)\right)$

$f\left(1,-\frac{\pi }{4}\right)=\tan \left(-\frac{\pi }{4}\right)=-\tan \left(\frac{\pi }{4}\right)$

Equation $4$

$f\left(1,-\frac{\pi }{4}\right)=-1$

$\left(\because \tan \left(\frac{\pi }{4}\right)=1\right)$

Step 3: Conclusion

Equations $2, 3$ and $4$ are substituted by equation $1$, we get,

$-\frac{\pi }{2}(x-1)+2\left(y+\frac{\pi }{4}\right)=z+1$

$-\frac{\pi }{2}x+\frac{\pi }{2}+2y+\frac{\pi }{2}=z+1$

width="199" height="41" style="max-width: none; vertical-align: -15px;" $-\frac{\pi }{2}x+2y-z=1-\frac{\pi }{2}-\frac{\pi }{2}$

$2$ is multiplied by two sides, we get,

$-\pi x+4y-2z=2-\pi -\pi$

$-\pi x+4y-2z=2-2\pi$

Finally, we get the result $\pi x-4y+2z=2\pi -2$.