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Consider the function f(x, y) = 4xy / (x² + 1)(y² + 1) on the intervals -2 ≤ x ≤ 2 and 0 ≤ y ≤ 3. Find a set of parametric equations of the normal line and an equation of the tangent plane to the surface at the point (1, 1, 1).


Answered 7 months ago
Answered 7 months ago
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Since f(x,y)=4xy(x2+1)(y2+1)f(x,y)=\dfrac{4xy}{(x^2+1)(y^2+1)}, we can written as:

F(x,y,z)=4xy(x2+1)(y2+1)z\begin{align*} F(x,y,z)=\dfrac{4xy}{(x^2+1)(y^2+1)}-z \end{align*}

The gradient\textbf{The gradient} of F(x,y,z)F(x,y,z) is given by:

F(x,y,z)=Fx(x,y,z)i+Fy(x,y,z)j+Fz(x,y,z)k\begin{align*} \color{#500000}\nabla F(x,y,z)=F_x(x,y,z)\mathbf{i}+F_y(x,y,z)\mathbf{j}+F_z(x,y,z)\mathbf{k}\tag*{\color{#500000}(1)} \end{align*}

To determine Fx(x,y,z)F_x(x,y,z), take the derivative of the function with respect to xx and treat y,zy,z as constants.

Fx(x,y,z)=x(4xy(x2+1)(y2+1)z)\begin{align*} F_x(x,y,z)=\frac{\partial \:}{\partial \:x}\left(\dfrac{4xy}{(x^2+1)(y^2+1)}-z\right) \end{align*}

Apply the Sum/Difference Rule:x(f±g)=fx±gx\quad \color{#500000}\dfrac{\partial \:}{\partial \:x}\left(f\pm g\right)=\dfrac{\partial f}{\partial x}\pm \dfrac{\partial g}{\partial x}.

Fx(x,y,z)=x(4xy(x2+1)(y2+1)z)=x(4xy(x2+1)(y2+1))x(z)\begin{align*} F_x(x,y,z)&=\frac{\partial \:}{\partial \:x}\left(\dfrac{4xy}{(x^2+1)(y^2+1)}-z\right)\\ &=\frac{\partial \:}{\partial \:x}\left(\frac{4xy}{\left(x^2+1\right)\left(y^2+1\right)}\right)-\frac{\partial \:}{\partial \:x}\left(z\right) \end{align*}

Derivative of a constant:x(a)=0\quad \color{#500000} \dfrac{\partial \:}{\partial \:x}\left(a\right)=0.

Fx(x,y,z)=x(4xy(x2+1)(y2+1))x(z)=x(4xy(x2+1)(y2+1))0=x(4xy(x2+1)(y2+1))\begin{align*} F_x(x,y,z)&=\frac{\partial \:}{\partial \:x}\left(\frac{4xy}{\left(x^2+1\right)\left(y^2+1\right)}\right)-\frac{\partial \:}{\partial \:x}\left(z\right)\\ &=\frac{\partial \:}{\partial \:x}\left(\frac{4xy}{\left(x^2+1\right)\left(y^2+1\right)}\right)-0\\ &=\frac{\partial \:}{\partial \:x}\left(\frac{4xy}{\left(x^2+1\right)\left(y^2+1\right)}\right) \end{align*}

Apply the constant multiple rule:x(af)=axf\quad \color{#500000}\dfrac{\partial \:}{\partial \:x}\left(a\cdot f\right)=a\cdot \dfrac{\partial \:}{\partial \:x}f with a=yy2+1a=\dfrac{y}{y^2+1} and f=4xx2+1f=\dfrac{4x}{x^2+1}.

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