Find the points on the cone z^2 = x^2 + y^2 that are closest to the point (2,2,0).

This question aims to explain the concepts of maxima and minima. Formulas to calculate the extreme values of the function. Further, it explains how to calculate the distance between the points.

In mathematics, the length of the line segment between the two points is the Euclidean distance between two points. The Pythagorean theorem is used to calculate the distance from the cartesian coordinates of the point. It is also called the Pythagorean distance.

The largest and smallest value of the function is called its maxima and minima respectively either for the entire domain or the given range. They are also called the extrema of the function.

Expert Answer

Let us assume the point $B(x,y,z)$ represents the point on the cone.

Finding the distance between the point $A(2,2,0)$ and the point $B(x,y,z)$:

Inserting the values in the distance formula:

$$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$$

$$d=\sqrt{(x-2{)}^2+(y-2{)}^2+(z-0{)}^2}$$

$$d=\sqrt{(x-2{)}^2+(y-2{)}^2+z^2}$$

Inserting the $z^2=x^2+y^2$ in the above equation:

$$d=\sqrt{(x-2{)}^2+(y-2{)}^2+x^2+y^2}$$

Squaring both sides:

$$d^2=(x-2{)}^2+(y-2{)}^2+x^2+y^2$$

If we minimize $d^2$, we minimize the distance $d$ between the points $A(2,2,0)$ and the point $B(x,y,z)$.

$$f^{\prime }=0$$

$${\displaystyle \frac{df}{dx}}={\displaystyle \frac{d}{dx}}(x-2{)}^2+(y-2{)}^2+x^2+y^2$$

$${\displaystyle \frac{df}{dx}}=2(x-2)+2x$$

Putting ${\displaystyle \frac{df}{dx}}$ equals to $0$ and solving for $x$:

$$2x–4+2x=0$$

$$4x=4$$

$$x=1$$

Similarly solving for $y$:

$${\displaystyle \frac{df}{dy}}={\displaystyle \frac{d}{dy}}(x-2{)}^2+(y-2{)}^2+x^2+y^2$$

$${\displaystyle \frac{df}{dy}}=2(y-2)+2y$$

Putting ${\displaystyle \frac{df}{dy}}$ equals to $0$ and solving for $y$:

$$2y–4+2y=0$$

$$4y=4$$

$$y=1$$

Now solving $z^2=x^2+y^2$ by inserting the above calculated values of $x$ and $y$.

$$z^2=1+1$$

$$z^2=2$$

$$z=\pm \sqrt{2}$$

Numerical Results

The points on the cone $z^2=x^2+y^2$ that are closest to the point $(2,2,0)$ are $(1,1,\sqrt{2})$ and $(1,1,-\sqrt{2})$.

Example

Find the points that are closest to the point $(4,2,0)$ on the cone $z^2=x^2+y^2$.

Assume the point $B(x,yz)$ to be the point on the cone.

The distance between the point $A(4,2,0)$ and the point $B(x,y,z)$ is:

$$d=\sqrt{(x-4{)}^2+(y-2{)}^2+(z-0{)}^2}$$

$$d=\sqrt{(x-4{)}^2+(y-2{)}^2+z^2}$$

Inserting $z^2$:

$$d=\sqrt{(x-4{)}^2+(y-2{)}^2+x^2+y^2}$$

$$d^2=(x-2{)}^2+(y-2{)}^2+x^2+y^2$$

Minimizing the distance $d$:

$$f^{\prime }=0$$

$${\displaystyle \frac{df}{dx}}={\displaystyle \frac{d}{dx}}(x-4{)}^2+(y-2{)}^2+x^2+y^2=0$$

$${\displaystyle \frac{df}{dx}}=2(x-4)+2x=0$$

$$2x-8+2x=0$$

$$4x=8$$

$$x=2$$

Similarly solving for $y$:

$${\displaystyle \frac{df}{dy}}={\displaystyle \frac{d}{dy}}(x-4{)}^2+(y-2{)}^2+x^2+y^2=0$$

$${\displaystyle \frac{df}{dy}}=2(y-2)+2y=0$$

$$2y-4+2y=0$$

$$4y=4$$

$$y=1$$

Now solving $z^2=x^2+y^2$ by inserting the above calculated values of $x$ and $y$.

$$z^2=2^2+1$$

$$z^2=5$$

$$z=\pm \sqrt{5}$$

Closest points are $(2,1,\sqrt{5})$ and $(2,1,-\sqrt{5})$

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