# Chapter 14 - Partial Derivatives - 14.6 Directional Derivatives and the Gradient Vector - 14.6 Exercises - Page 999: 56

Two surfaces are tangent to each other at the point $(1,1,2)$. This means that they have a common tangent plane at the point.

#### Work Step by Step

Our aim is to determine the tangent plane equation for an ellipsoid. The general form is: $(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$ ...(1) From the given data, we have $f(x,y,z)=(1,1,2)$ Equation (1), becomes: Thus, $(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$ or, $(x-x_0)(6x)+(y-y_0)(4y)+(z-z_0)(-2z)=0$ or, $(x-1)(6)+(y-1)(4)+(z+1)(-4)=0$ or, $6x-6+4y-4-4z-4=0$ thus, $6x+4y-4z=18$ ....(a) Our aim is to determine the tangent plane equation for a sphere. $(x_2-x_1)f_x(x_1,y_1,z_1)+(y_2-y_1)f_y(x_1,y_1,z_1)+(z_2-z_1)f_x(x_1,y_1,z_1)=0$ From the given data, we have $f(x,y,z)=(1,1,2)$$(x-x_0)(2x-8)+(y-y_0)(2y-6)+(z-z_0)(2z-8)=0$ Thus,$(x-1)(6)+(y-1)(4)+(z+1)(-4)=0$ or, $6x-6+4y-4-4z-4=0$ or, $6x+4y-4z=18$ ...(b) Our equations (a) and (b) tells us that the two surfaces can be tangent to each other at the point $(1,1,2)$. Also, this means that they would include a common tangent plane at the point$(1,1,2)$.

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