Q 41. - Calculus

Question

In Exercises $37-42$, use the partial derivatives of $f\left(x,y\right)=x\sin y$ and the point $\left(2,\frac{\pi }{3}\right)$ specified to

$\left(a\right)$ find the equation of the line tangent to the surface defined by the function in the $x$ direction,

$\left(b\right)$ find the equation of the line tangent to the surface defined by the function in the $y$ direction, and

$\left(c\right)$ find the equation of the plane containing the lines you found in parts $\left(a\right)$ and $\left(b\right)$.

Step-by-Step Solution

Answer

Part $\left(a\right)$, The equation of the line tangent to the surface defined by the function in the $x$ direction is

$x=2+t, y=\frac{\pi }{3}, z=\frac{3\sqrt{3}}{2}t$

Part $\left(b\right)$, The equation of the line tangent to the surface defined by the function in the $y$ direction is

$x=2, y=\frac{\pi }{3}+t, z=\sqrt{3}+t$

Part $\left(c\right)$, The equation of the plane containing the lines you found in parts $\left(a\right)$ and $\left(b\right)$ is

$3\sqrt{3}x+6y-6z=2\pi$

Step 1. Explanation of part a , Finding equation of tangent in x direction

The line tangent to the surface at $\left(a,b,f\left(a,b\right)\right)$ in the $x$ direction is given by the parametric equation

$x=a+t, y=b, z=f\left(a,b\right)+f_x\left(a,b\right)t$

Now we have function $f\left(x,y\right)=x\sin y$ and point $\left(2,\frac{\pi }{3}\right)$

So $a=2, b=\frac{\pi }{3}, f\left(a,b\right)=\sqrt{3}, f_x\left(a,b\right)=\frac{\sqrt{3}}{2}$

Therefore, equation of tangent in $x$ direction is

$x=2+t, y=\frac{\pi }{3}, z=\frac{3\sqrt{3}}{2}t$

Step 2. Explanation of part b , Finding equation of tangent in y direction

The line tangent to the surface at $\left(a,b,f\left(a,b\right)\right)$ in the $y$ direction is given by the parametric equation

$x=a, y=b+t, z=f\left(a,b\right)+f_y\left(a,b\right)t$

Now we have function $f\left(x,y\right)=x\sin y$ and point $\left(2,\frac{\pi }{3}\right)$

So $x=2, y=\frac{\pi }{3}, f\left(a,b\right)=\sqrt{3}, f_y\left(a,b\right)=1$

Therefore, equation of tangent in $y$ direction is

$x=2, y=\frac{\pi }{3}+t, z=\sqrt{3}+t$

Step 3. Explanation of part c , Finding equation of plane

The equation of plane containing the given lines and point is  

$f_x\left(a,b\right)\left(x-a\right)+f_y\left(a,b\right)\left( y-b\right)=z-f\left(a,b\right)$

$\Rightarrow \frac{\sqrt{3}}{2}\left(x-2\right)+1\left(y-\frac{\pi }{3}\right)=z-\sqrt{3}$

$\Rightarrow \frac{\sqrt{3}}{2}x+y-z=\frac{\pi }{3}$

$\Rightarrow 3\sqrt{3}x+6y-6z=2\pi$