# Difference between a Gradient and Tangent

I am unable to understand the fundamental difference between a Gradient vector and a Tangent vector. I need to understand the geometrical difference between the both.

By Gradient I mean a vector $\nabla F(X)$ , where $X \in [X_1 X_2\cdots X_n]^T$

Note: I saw similar questions on "Difference between a Slope and Gradient" but the answers didn't help me much.

Appreciate any effort.

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@JesseMadnick : Please refer the edit –  Jugesh Sundram Jan 30 at 22:51
The gradient is a vector associated with a scalar field--a real-valued function of several real variables. Usually, a tangent vector is associated with a curve--a vector-valued function of a single variable. Is this the kind of tangent vector you're referring to? –  Muphrid Jan 30 at 22:55
The gradient of a function $(x_1,x_2,\ldots,x_n)\mapsto y$ is the vector $\left(\dfrac{\partial y}{\partial x_1},\dfrac{\partial y}{\partial x_2},\ldots,\dfrac{\partial y}{\partial x_n}\right)$. The tangent to a curve $x\mapsto(y_1,y_2,\ldots,y_n)$ is the vector $\left(\dfrac{\mathrm dy_1}{\mathrm dx},\dfrac{\mathrm dy_2}{\mathrm dx},\ldots,\dfrac{\mathrm dy_n}{\mathrm dx}\right)$. What's the problem? :) –  Rahul Narain Jan 30 at 23:02
So... One is $\partial \phi / \partial x_i$, the other is $d\vec{f(x)}/dx$ (ninja'd by $\mathbb{R}^n$) –  NeuroFuzzy Jan 30 at 23:03
So basically, the Gradient vector is applicable only in a scalar field and the Tangent vector belongs to the vector. It was quite foolish of me to mesh the 2 concepts together. I was under the impression we can apply Tangent vectors to scalar field. But it doesn't make sense at all. Thanks all! –  Jugesh Sundram Jan 30 at 23:10
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## 2 Answers

I suppose the question has been answered in the comments.

The gradient of a function $(x_1,x_2,\ldots,x_n)\mapsto y$ is the vector $\left(\dfrac{\partial y}{\partial x_1},\dfrac{\partial y}{\partial x_2},\ldots,\dfrac{\partial y}{\partial x_n}\right)$.

The tangent to a curve $x\mapsto(y_1,y_2,\ldots,y_m)$ is the vector $\left(\dfrac{\mathrm dy_1}{\mathrm dx},\dfrac{\mathrm dy_2}{\mathrm dx},\ldots,\dfrac{\mathrm dy_n}{\mathrm dx}\right)$.

Both can be thought of as special cases of the Jacobian of a vector-valued multivariate function $(x_1,x_2,\ldots,x_n)\mapsto(y_1,y_2,\ldots,y_m)$, which is the matrix $$\begin{pmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} & \cdots & \frac{\partial y_1}{\partial x_n}\\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} & \cdots & \frac{\partial y_2}{\partial x_n}\\ \vdots & \vdots & \ddots & \vdots \\ \frac{\partial y_m}{\partial x_1} & \frac{\partial y_m}{\partial x_2} & \cdots & \frac{\partial y_m}{\partial x_n}\\ \end{pmatrix}$$

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Say you are standing on the side of a hill. Imagine somewhere beneath the hill, there is a flat $x,y$ plane that you can use to determine your position. Let's say $+x$ is east and $+y$ is north.

If the hill is smooth, then the height of the hill above this plane is some continuous function $f(x,y)$.

The gradient of $f$ at any point tells you which direction is the steepest from that point and how steep it is. To find the direction of the gradient of $f$ where you are standing, decide which direction is the steepest. The answer could be "north" or "30 degrees west of south". There is no vertical component to the gradient, it is telling you a direction with respect to the $x,y$ plane which is your reference. The magnitude of the gradient will be the slope of the hill in that direction.

The tangent plane is the plane that best approximates the shape of the hill where you are standing. The hill may be curved if you look at it from a distance, but maybe directly beneath your feet it is flat enough to set a pizza box down and have it be flush with the ground. The plane that the bottom of the pizza box defines would, roughly, be the "tangent" plane.

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thanks for the visual explanation.. it helped me understand the difference :) –  Jugesh Sundram Jan 31 at 19:20
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