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Problem of the Week #88 - December 2nd, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: Suppose $f$ is a differentiable function of one variable. Show that all tangent planes to the surface $z=xf(y/x)$ intersect in a common point.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
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This week's problem was correctly answered by MarkFL. You can find his solution below.

We are given:

\(\displaystyle z(x,y)=xf\left(\frac{y}{x} \right)\)

Let us the define:

\(\displaystyle F(x,y,z)=xf\left(\frac{y}{x} \right)-z\)

Thus, a point $\left(x_0,y_0,z_0 \right)$ is on the graph of \(\displaystyle z(x,y)=xf\left(\frac{y}{x} \right)\) iff it is also on the level surface $F(x,y,z)=0$. This follows from \(\displaystyle F\left(x_0,y_0,z_0 \right)=x_0f\left(\frac{y_0}{x_0} \right)-z_0=0\).

Next, we may utilize the following theorem:

Equation of Tangent Plane

Let $P\left(x_0,y_0,z_0 \right)$ be a point on the graph of $F(x,y,z)=c$ where $\nabla F\ne0$. Then an equation of the tangent plane at $P$ is:

\(\displaystyle F_x\left(x_0,y_0,z_0 \right)\left(x-x_0 \right)+F_y\left(x_0,y_0,z_0 \right)\left(y-y_0 \right)+F_z\left(x_0,y_0,z_0 \right)\left(z-z_0 \right)=0\)

Computing the required partials, we find:

\(\displaystyle F_x(x,y,z)=xf'\left(\frac{y}{x} \right)\left(-\frac{y}{x^2} \right)+f\left(\frac{y}{x} \right)=f\left(\frac{y}{x} \right)-\frac{y}{x}f'\left(\frac{y}{x} \right)\)

\(\displaystyle F_y(x,y,z)=xf'\left(\frac{y}{x} \right)\left(\frac{1}{x} \right)=f'\left(\frac{y}{x} \right)\)

\(\displaystyle F_z(x,y,z)=-1\)

Using the above theorem, we then find the equation of the tangent plant at $P$ is:

\(\displaystyle \left(f\left(\frac{y_0}{x_0} \right)-\frac{y_0}{x_0}f'\left(\frac{y_0}{x_0} \right) \right)\left(x-x_0 \right)+\left(f'\left(\frac{y_0}{x_0} \right) \right)\left(y-y_0 \right)+(-1)\left(z-z_0 \right)=0\)

Distributing and rearranging, we obtain:

\(\displaystyle ]\left(f\left(\frac{y_0}{x_0} \right)-\frac{y_0}{x_0}f'\left(\frac{y_0}{x_0} \right) \right)x+\left(f'\left(\frac{y_0}{x_0} \right) \right)y-z-\left(x_0f\left(\frac{y_0}{x_0} \right)-z_0 \right)=0\)

Now, given \(\displaystyle F\left(x_0,y_0,z_0 \right)=x_0f\left(\frac{y_0}{x_0} \right)-z_0=0\), this reduces to:

\(\displaystyle ]\left(f\left(\frac{y_0}{x_0} \right)-\frac{y_0}{x_0}f'\left(\frac{y_0}{x_0} \right) \right)x+\left(f'\left(\frac{y_0}{x_0} \right) \right)y-z=0\)

Observing that all planes given by $ax+by+cz=0$ pass through the origin, we therefore find that all of the planes tangent to $z$ must have the origin as a common point.
 
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