Question

You are climbing a mountain by the steepest route at a slope of 10∘ when you come upon a trail branching off at a 45∘ angle from yours. What is the angle of ascent of the branch trail?

Answer 1

try 7.11 degrees

Answer 2

@ganeshie8 7.11 is correct, why?

Answer 3

im not good with these, some what tricky to explain what i did.. wait

Answer 4

imagine you're at origin when you saw another trial 45 degrees from your trail

Answer 5

Lets say \(z = f(x,y)\) is the surface of mountain.

then the directional derivative along a specific unit vector gives the slope (ascent) along that direction, yes ?

Answer 6

@ganeshie8 ty for the reply, sorry for delayed response I was having internet issues.

Answer 7

still am

Answer 8

ive seen the answer is arctan(cos(45)tan(10)) but dont really get why

Answer 9

Ohhk... recall below stuff :
\[\text{slope}= \tan(\theta) = \dfrac{dy}{dx}\]

gradient vector points in your direction because you're climbing along the steepest trail.

since the angle along your trail is fixed at \(10^{\circ}\), would you agree that the x component of gradient vector is \(\tan(10^{\circ})\) ?

\[\nabla f = \langle \tan{10^{\circ}}, ~y \rangle \]

Answer 10

yep that makes sense other than i would think its the y component not the x component

Answer 11

or maybe the z component?

Answer 12

it has to be the x component because we have assumed your path is along x direction

Answer 13

oh i see, from your drawing, ok

Answer 14

two variables are enough to represent any motion on a surface

Answer 15

ah ok

Answer 16

the z value is just the output of the function of the surface

Answer 17

z = f(x,y) is a function which spits out the elevation of mountain (surface) based on the coordinates in xy plane

Answer 18

you're right

Answer 19

k

Answer 20

i get confused sometimes with this stuff

Answer 21

me too haha

Answer 22

so far we have gradient vector represented as below
\[\nabla f = \langle \tan{10^{\circ}}, ~y \rangle\]

Answer 23

k

Answer 24

the y component of gradient must be 0 because the gradient is pointing along x direction at (0, 0) :

\[\nabla f(0,0) = \langle \tan{10^{\circ}}, ~0 \rangle\]

Answer 25

k

Answer 26

Now that you know the gradient at (0,0), you're ready to find the directional derivative along 45 degrees trail.

just find the unit vector and multiply it by the gradient

Answer 27

unit vector along 45 degree trail = \(\langle \cos(45), ~\sin(45) \rangle \)

Answer 28

so the directional derivative along 45 degree trail would be
\[ \dfrac{df}{dx}\Bigg|_{45^{
\circ}\text{ branch trail} } = \langle \tan{10^{\circ}}, ~0 \rangle \cdot \langle \cos(45), ~\sin(45) \rangle = \tan{10^{\circ}} \cos(45)\]

Answer 29

very nice, ty so much, makes sense now :)

Answer 30

for angle, just set that derivative equal to \(\tan(\theta)\) and solve \(\theta\) :

\[\tan(\theta) = \tan{10^{\circ}} \cos(45)\]

\[\theta = \arctan(\tan{10^{\circ}} \cos(45))\]

Answer 31

oops, i spoke too soon, why do we set derivative = tan(theta)?

Answer 32

ah yes, ok, got it, opposite/adj, dy/dx