Q. 18 - Calculus

Question

Let $$u_{1}$$, $$u_{2}$$, and $$u_{3}$$ be three unit vectors in $$R^{3}$$ that cannot be put into the same plane. If the function $$f(x, y,z)$$ is differentiable at a point $$(a, b,c)$$, explain why the tangent lines to the graph of $$f$$ at $$(a, b,c)$$ in the $$u_{1}$$, $$u_{2}$$, and $$u_{3}$$ directions are sufficient to determine the tangent hyperplane to the surface.

Step-by-Step Solution

Answer

If the function $$f(x, y,z)$$ is differentiable at a point $$(a, b,c)$$, then the tangent lines to the graph of $$f$$ at $$(a, b,c)$$ in the $$u_{1}$$, $$u_{2}$$, and $$u_{3}$$ directions are sufficient to determine the tangent hyperplane to the surface since they all lie in the same hyperplane.

Step 1. Given Information

• Let $$u_{1}$$, $$u_{2}$$, and $$u_{3}$$ be three unit vectors in $$R^{3}$$ that cannot be put into the same plane. 
• The function, $$f(x, y,z)$$ is differentiable at a point $$(a, b,c)$$.

Step 2. Explanation

It is given that the function $$f(x, y,z)$$ is differentiable at a point $$(a, b,c)$$. 

$$\implies$$ All the lines tangent to the graph of the function, $$f$$ at the point $$(a,b,c)$$ lie in the same hyperplane.

So, we can determine the equation of the hyperplane using any three non-coplanar lines in that same hyperplane.

Here, $$u_{1}$$, $$u_{2}$$, and $$u_{3}$$ be three unit vectors in $$R^{3}$$ that cannot be put into the same plane. or are non-coplanar vectors.

Hence, if the function $$f(x, y,z)$$ is differentiable at a point $$(a, b,c)$$, then the tangent lines to the graph of $$f$$ at $$(a, b,c)$$ in the $$u_{1}$$, $$u_{2}$$, and $$u_{3}$$ directions are sufficient to determine the tangent hyperplane to the surface.