# Environmental Science – Grad, Div and Curl (1/3)

Narrator: (Francesca Hunt)
Skiers often zig-zag across a slope rather than going straight down it; they have to take a path that’s not too steep, so they can control their speed. Even expert skiers like Carl can’t ski down slopes of more than 25 to 30 degrees! Carl: Going straight down the slope is the quickest way of getting down the slope. Alternatively you can go traverse which means skiing across the slope, that way you control your descent and you get down at a more controlled speed and more safely. Narrator: But what do we mean by ‘the steepness of a path’ on a hillside? How can we describe the steepness of the path that goes straight up or down the slope? And how is it related to the steepness of any other chosen path? To answer those questions we’ll need to build a model of a hillside. We’ll model the surface of the hillside as a function of two variables. The plane you can see at the base is the x-y plane. The height z of any point on the surface itself is a unique scalar value, f(x,y). So the height z is a scalar field. In fact here f(x,y) is equal to sin x cos y, but that’s not important. To find the slope of the steepest path at
any point we’ll use the tangent plane. The direction of the line of steepest upward slope on the tangent plane is shown by the arrow. No matter where we are on the surface, the arrow is always at right angles to the contour line passing through that point. So, if we look at the projection on the x – y plane, at any point we can define a vector, in the x-y plane, called grad f. The direction of grad f specifies the direction of the steepest slope while the magnitude of grad f is simply the magnitude of the steepest slope. You’ll see that grad f is given by
this definition involving the partial derivatives of f. The gradient vector field for the whole surface looks like this. However, as we’ve heard, skiers don’t usually take the steepest path
down a hillside. So how can we find the slope of a path in any other direction? It turns out that provided we know grad, it’s not too hard to find. Let’s go back and look at the hillside
again. This is Brent Knoll in Somerset. In fact to find the slope of any path on this hillside we don’t need a mathematical model; we can estimate the slope from a map! Each of the contours on this map represents a constant height above sea level. Earlier we saw that the magnitude and direction of the steepest slope of a path at any point are given by the gradient vector grad f. So let’s find the gradient vector at this point A. We’re assuming that the ground
between these two contour lines is in a plane. Now we know that the direction of grad f is perpendicular to the contour line through A, and so will be parallel to a unit vector n uphill. The labelling on the contours shows that the rise in height along a path from the point A to the point B along the direction of grad is 10 metres. And as it’s a map, that rise in height corresponds to going a horizontal distance AB. Using the scale on this map that works
out to be 30 metres. So grad f is ‘rise over run’; 10, divided by 30, in the direction n. In other words grad f is 1/3 n. So now we know the magnitude and direction of the gradient vector. But what’s the slope of any other path? Say this one, passing through the points A and C? The direction of the path is parallel to this
unit vector, u. The slope of the path through AC is given by the same rise of 10 metres marked on the contour lines but divided this time by the horizontal distance corresponding to AC; that’s 50 metres. So the slope of AC is 1/5. But all we’ve done so far is find the slope from the map. How does all this relate to the gradient vector grad? Well, remember the contour lines
are parallel, and so ABC is a right angled triangle. So AB is equal to AC times cos – which here means that 30=50 cos . Also u is a unit vector in the direction of AC, so cos is the dot product of the two unit vectors u and n. Putting all this together, we get the result that the slope of the path AC is equal to
u. grad f. This is a general result, showing how the
slope of any path is related to grad f. In fact you’ve already seen another example of this result; the dot product of perpendicular vectors is zero, so the gradient vector must be perpendicular to the contour line at any point. So if skiers knew f, and the direction of any proposed path, they could work out the slope of the path by taking the dot product of the unit vector in the direction of their proposed path with grad f!

• Tolson Winters

Thanks! You do a great job teaching with some fun examples you don't always see in just black & white textbooks 😉 The visual aid is admirable

• Prasan Dutt

Great source of information. I got motivation for reading vector calculus.

• ali hamwi

very good…..but it is difficult….the example on 23 is simpler

• morizl

thank you, the video really helped me understand the definition of grad a lot!

but i have to say: you call this skiing?! haha are you kidding me? and expert skiiers can't go down steeper than 30°? i am from austria and i can say 45° are no problem for a lot of people (for me under 30° is broeing) and expert skiiers can go down more than 70° ( and no im not confuseing with %, if you don't beleve me look at the last 3 min of this St. Anton am Arlberg – Freeride Deluxe 2012 ) and by the way skiing is a SNOW-sprt 😉

• Stephen Rayner

rotating to fast making me feel dizzy…

• Paul Miller

5:13–5:30 not a very 'accurate' diagram in that case. It looks more like AC = 30/cos(1/2)m – or about 33.5m. That's not very much more than AB in fact.

• dlwatib

I think you left out a few steps of explanation. At 2:48 you give an equation that you don't derive from anything and don't use again. Why did you bother to give it? At 4:41 you state without any justification or explanation that the magnitude if grad f is rise over run. How does this relate to the previous formula, please?

• Chris Norwall

i appreciate this visual demonstration, helps to understand the concept

• oussama Ali

thanks again

• Daan G

So helpful. Thank you

• Ramin Golbang

Excellent!

• Aakash Pathak

Absolutely brilliant video!! I really appreciate such videos a lot because they bring reality into the black and white world of the book. Even a 8 year old would believe that if he tries he can understand the concept because its seems so easy, however, it is no argument that he would never care to open the black and white book. This video reminds me of the 8 year old me, trying to understand everything. I learn better when away from the mundane, the black and white!

• Tasnimul Hoque

thanks lot…it is too much helpful and easy…. thanks
]

• vinay kumar

excellent explnation

• Riken Maharjan

In a 3d mountain graph, is the gradient in the xy plane or in the 3d space. Please anyone help me. The video I watched prior to this said it was in xy plane.If it was in xy plane, what does n unit vector direction is in. Isn't n unit vextor just in xyplane: But now here, they say Grad F is hypotenuse where perpendicular length =10 and base length=30 @4:45, which means gradient is not in xy plane.
Video I watched prior to this: https://www.youtube.com/watch?v=GkB4vW16QHI
Anyone . PLeaseeeee…… help..

• Wardell Lindsay

excellent explanations!
The Universe is NOT expanding. Dark Energy is cP. Gravity energy is
W=[c,V][,P]=[-vp,cP]=
[-mGM/r, cP]
Planck's constant=Mq=zq^2=h where z=375 Ohm and h=2/3E-33.
z is free space impedance.

• jamesmasonic

i dont understand

• Govindaraj P

Don't understand…,

• Govindaraj P

please explain another real-time example

• Akito

Their acting is so cringe.

• MuscleSheriff

0:11 it's because they're scared

XD

• suse me

Sir if we through the envolep in flowing water then why he up and down in water

• Khaled Farrag

how apply online for mathematics and science
major degree

• Hsi S

Dang , nice effort but its dumb fcked my mind. I lost in the mid way.