## 17Calculus Partial Derivatives - The Gradient Vector

##### 17Calculus

Gradient Vector

The directional derivative and gradient are very closely related. On this page we give an overview of both but discuss only the gradient in detail. On the directional derivative page you will find a more complete discussion of the directional derivative. However, you need to understand the gradient first.

Topics You Need To Understand For This Page

Overview of the Gradient and the Directional Derivative

Directional Derivative - - You know from single variable calculus that the derivative is just the slope of the tangent line. If you are given a point, you can find the slope of the specific tangent line at that specific point. Similarly, the directional derivative is the slope of a tangent line but it applies to three dimensions where you are given a point and a unit vector. This unit vector tells you which specific tangent line you are talking about at that point. In summary, a directional derivative is the slope of a tangent line in a specific direction.

Gradient - - The gradient is a vector (which is why its also called the gradient vector) calculated using the partial derivative. It contains information about the slope at every point. We can use the gradient to get the directional derivative.

Remember that in multi-variable calculus most concepts use special notation. The directional derivative and gradient are no exception.

Gradient

$$\displaystyle{ \nabla g(x,y) = \frac{\partial g}{\partial x} \hat{i} + \frac{\partial g}{ \partial y}\hat{j} }$$ where $$\displaystyle{ \nabla }$$ is the 'del' operator $$\displaystyle{ \nabla = \frac{\partial }{\partial x}\hat{i} + \frac{\partial }{ \partial y}\hat{j} }$$

The gradient can also be indicated by grad(g) and can be called the grad of g. Notice that we do not use vector notation on $$\nabla g(x,y)$$. We could write $$\vec{ \nabla } g(x,y)$$ to clearly indicate that the result is a vector and some instructors do this. However, many of them don't and we won't either. So you just need to remember that $$\nabla g(x,y)$$ is a vector (or, more precisely, a vector field).

So you may be asking yourself, what does the gradient actually mean? What does it look like? Well, here is a great video that will give you a more intuitive understanding of the gradient vector.

### Trefor Bazett - Geometric Meaning of the Gradient Vector

video by Trefor Bazett

Functions of More Than Two Variables

You can easily extend the definition of the gradient and directional derivative to functions of 3 or more variables. However, as you increase the number of variables, you lose the physical interpretation of slope. For three dimensions, the gradient looks like you would expect.

 $$\displaystyle{ \nabla h(x,y,z) = \frac{\partial h}{\partial x} \hat{i} + \frac{\partial h}{ \partial y}\hat{j} + \frac{\partial h}{ \partial z}\hat{k} }$$

Before going on to the directional derivative, let's make sure we understand the gradient by working a few practice problems.

Practice

Unless otherwise instructed, find the gradient vector of these functions.

$$f(x,y) = x\sin y + \cos y$$

Problem Statement

Find the gradient of $$f(x,y) = x\sin y + \cos y$$

Final Answer

$$\nabla f(x,y) = (\sin y) \vhat{i} + (x\cos y - \sin y)\vhat{j}$$

Problem Statement

Find the gradient of $$f(x,y) = x\sin y + \cos y$$

Solution

 $$\displaystyle{ \nabla f(x,y) = \frac{\partial f}{\partial x} \vhat{i} + \frac{\partial f}{\partial y} \vhat{j} }$$ $$= (\sin y + 0) \vhat{i} + (x\cos y - \sin y)\vhat{j}$$ $$= (\sin y) \vhat{i} + (x\cos y - \sin y) \vhat{j}$$

Final Answer

$$\nabla f(x,y) = (\sin y) \vhat{i} + (x\cos y - \sin y)\vhat{j}$$

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If $$f(x,y,z) = x^3+y+z^2$$, calculate $$\nabla f$$ and $$\nabla f(1,2,3)$$.

Problem Statement

If $$f(x,y,z) = x^3+y+z^2$$, calculate $$\nabla f$$ and $$\nabla f(1,2,3)$$.

Solution

### Dr Chris Tisdell - 804 video solution

video by Dr Chris Tisdell

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Calculate $$\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }$$ at $$(\tau, \kappa) = (1,5)$$

Problem Statement

Calculate $$\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }$$ at $$(\tau, \kappa) = (1,5)$$

Hint

This problem uses unusual variables $$\tau$$ and $$\kappa$$. However, they work just as if you had t and x.

Problem Statement

Calculate $$\displaystyle{ \nabla \left( \frac{\kappa^2 \arctan \tau}{\kappa-\tau^3} \right) }$$ at $$(\tau, \kappa) = (1,5)$$

Hint

This problem uses unusual variables $$\tau$$ and $$\kappa$$. However, they work just as if you had t and x.

Solution

### World Wide Center of Mathematics - 2293 video solution

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Calculate the gradient vector field of $$f(x,y) = \ln(2x+5y)$$.

Problem Statement

Calculate the gradient vector field of $$f(x,y) = \ln(2x+5y)$$.

Solution

### MIP4U - 2294 video solution

video by MIP4U

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Given $$f(x,y) = xy$$, determine $$\nabla f(-3,2)$$.

Problem Statement

Given $$f(x,y) = xy$$, determine $$\nabla f(-3,2)$$.

Solution

### MIP4U - 2295 video solution

video by MIP4U

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$$f(x,y,z) = z e^{-xy}$$

Problem Statement

Find the gradient of $$f(x,y,z) = z e^{-xy}$$

Final Answer

$$\nabla f(x,y,z) = -yze^{-xy} \vhat{i} - xz e^{-xy} \vhat{j} + e^{-xy} \vhat{k}$$

Problem Statement

Find the gradient of $$f(x,y,z) = z e^{-xy}$$

Solution

 $$\displaystyle{ \nabla f(x,y,z) = \frac{\partial f}{\partial x} \vhat{i} + \frac{\partial f}{\partial y} \vhat{j} + \frac{\partial f}{\partial z} \vhat{k} }$$ $$\displaystyle{ z(-y)e^{-xy} \vhat{i} + z (-x) e^{-xy} \vhat{j} + e^{-xy} \vhat{k} }$$

Final Answer

$$\nabla f(x,y,z) = -yze^{-xy} \vhat{i} - xz e^{-xy} \vhat{j} + e^{-xy} \vhat{k}$$

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$$f(x,y) = 3x + 7y$$

Problem Statement

Find the gradient vector of $$f(x,y) = 3x + 7y$$

Solution

### Thomas Wernau - 4340 video solution

video by Thomas Wernau

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$$f(x,y) = \sin(xy)$$

Problem Statement

Find the gradient vector of $$f(x,y) = \sin(xy)$$

Solution

### Thomas Wernau - 4341 video solution

video by Thomas Wernau

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$$f(x,y) = \ln x + \arctan(xy)$$

Problem Statement

Find the gradient vector of $$f(x,y) = \ln x + \arctan(xy)$$

Solution

### Thomas Wernau - 4342 video solution

video by Thomas Wernau

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$$f(x,y) = xyz^{-1}$$

Problem Statement

Find the gradient vector of $$f(x,y) = xyz^{-1}$$

Solution

### Thomas Wernau - 4343 video solution

video by Thomas Wernau

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