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Question: How do you sketch \(f\left( x,y \right)=\ln \left( {{x}^{2}}+{{y}^{2}} \right)\)?...

How do you sketch f(x,y)=ln⁡(x2+y2)f\left( x,y \right)=\ln \left( {{x}^{2}}+{{y}^{2}} \right)?

Explanation

Solution

The equation f(x,y)=ln⁡(x2+y2)f\left( x,y \right)=\ln \left( {{x}^{2}}+{{y}^{2}} \right) involves two independent variables, x and y. For sketching it, we need equate the given function to the variable z to get the equation z=ln⁡(x2+y2)z=\ln \left( {{x}^{2}}+{{y}^{2}} \right). For plotting it, we must first convert it into the polar form by substituting x=rcos⁡θx=r\cos \theta and y=rsin⁡θy=r\sin \theta . Then the given equation will get reduced to z=2ln⁡(r)z=2\ln \left( r \right). Then we have to sketch the two dimensional graph of the equation z=ln⁡(x)z=\ln \left( x \right) and finally rotate it about the z-axis. The required graph will be obtained as the surface of revolution.

Complete step by step solution:
The given function is
f(x,y)=ln⁡(x2+y2)f\left( x,y \right)=\ln \left( {{x}^{2}}+{{y}^{2}} \right)
As can be observed above, that ff is a function of the independent variables x and y. So we must have a dependent variable z which must be equated to ff to get the equation
⇒z=ln⁡(x2+y2)\Rightarrow z=\ln \left( {{x}^{2}}+{{y}^{2}} \right)
If we substitute x=rcos⁡θx=r\cos \theta and y=rsin⁡θy=r\sin \theta in the above equation, we get
⇒z=ln⁡((rcos⁡θ)2+(rsin⁡θ)2) ⇒z=ln⁡(r2cos⁡2θ+r2sin⁡2θ) \begin{aligned} & \Rightarrow z=\ln \left( {{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}} \right) \\\ & \Rightarrow z=\ln \left( {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta \right) \\\ \end{aligned}
Taking r2{{r}^{2}} common we get
⇒z=ln⁡(r2(cos⁡2θ+sin⁡2θ))\Rightarrow z=\ln \left( {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right) \right)
We know that cos⁡2θ+sin⁡2θ=1{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1. Substituting this above, we get
⇒z=ln⁡(r2)\Rightarrow z=\ln \left( {{r}^{2}} \right)
Now, we know from the properties of logarithm that ln⁡(am)=mln⁡(a)\ln \left( {{a}^{m}} \right)=m\ln \left( a \right). So the above equation can be written as
⇒z=2ln⁡(r)\Rightarrow z=2\ln \left( r \right)
If we replace rr by xx in the above equation, we will get
⇒z=2ln⁡(x)\Rightarrow z=2\ln \left( x \right)
The graph of the above equation in the X-Z plane can be plotted as

Now, on revolving the above graph about the z-axis, we will obtain the graph of the given function as a surface of revolution as shown below.

Hence, we have sketched the graph of the given function.

Note: The graph obtained in the above solution, as we can observe, is a three dimensional graph as it involves three variables, namely x, y, and z. These three dimensional graphs cannot be obtained using the simple derivatives and coordinates concepts. To obtain such graphs, we must be familiar with the advanced mathematics concepts such as the solids of revolution etc.