vector field questions (pure 6 ish)

Would be really grateful for any help at all with these- i have attempted some parts and put down my ideas but they may not be right:

1. for function f(x,y,z)= ln(x^2+y^2) + z find grad f

for this i get 2x/(x^2+y^2) i + 2y/(x^2+y^2) j + k

now consider a cylinder of radius 5 whose axis is centered along the z-axis

i) what is the rate of change of f(x,y,z) in the direction normal to the cylinder at the point (3,-4,4)?

is this the differential of grad f wrt something? Can i just sub in the points to the grad f above or do i need to do something else to it like differentiate again?

ii) What is the rate of change if f(x,y,z) in the direction m= i + 2j at the same point, where i,j are unit vectors parallel to the x,y axes respectively?

again i think i need grad f and differentiate this using the direction somehow then sub in the points but not sure how.

6.i) Obtain the equation of the plane that is tangent to the surface
z= 3(x^2)ysin((pi*x)/2) at the point x=y=1.

for this to find grad f i get 6xy(pi/2)cos((pi*x)/2) i + 3x^2 sin ((pi*x)/2) j

so using the point (1,1,0) this would give grad f as (0,3,0) at this point

so is the equation of the plane (r-(1,1,o))dot(0,3,0)=0 ?

ii)Take north to be the direction (0,1,0) and east to be the direction (1,0,0), in which direction will a marble roll if placed on the surface at x=1 y=(1/2)?

assuming these points have to be subs into grad f i think it gives me (0,3,0) so does that mean the marble will roll in the y direction i.e. north?

thanks v much, even if its just checking some of the answers i already have x

Reply 1

k@tie

10

I think for Question 1 - both parts, use the result
df/ds = u<hat>.grad f
So work out grad f, and substitute your point A into it. Work out u<hat> as the direction (normal to the cylinder at A) (so, (0,0,4) - (3,-4,4)) divided by the magnitude of this vector - then dot the product. For the second part do the same thing, but use the direction they give you for the u<hat>

Reply 2

Jonny W

8

Jitterbug

Would be really grateful for any help at all with these- i have attempted some parts and put down my ideas but they may not be right:

1. for function f(x,y,z)= ln(x^2+y^2) + z find grad f

for this i get 2x/(x^2+y^2) i + 2y/(x^2+y^2) j + k

now consider a cylinder of radius 5 whose axis is centered along the z-axis

i) what is the rate of change of f(x,y,z) in the direction normal to the cylinder at the point (3,-4,4)?

I agree with your calculation of

\nabla

f.

The outward unit normal to the cylinder at (3, -4, 4) is (1/5)(3, -4, 0).

We want the derivative wrt s of

f((3, -4, 4) + s(1/5)(3, -4, 0))

evaluated at s = 0. That derivative is

(1/5)(3, -4, 0).

\nabla

f(3, -4, 4)
= (1/5)(3, -4, 0).(6/25, -8/25, 1)
= 2/5

Jitterbug

ii) What is the rate of change if f(x,y,z) in the direction m= i + 2j at the same point, where i,j are unit vectors parallel to the x,y axes respectively?
We want the derivative wrt s of

f((3, -4, 4) + s(1/sqrt(5))(1, 2, 0))

evaluated at s = 0. That derivative is

(1/sqrt(5))(1, 2, 0).

\nabla

f(3, -4, 4)
= (1/sqrt(5))(1, 2, 0).(6/25, -8/25, 1)
= -(1/sqrt(5))10/25
= -2/(5sqrt(5))

Jitterbug

6.i) Obtain the equation of the plane that is tangent to the surface
z= 3(x^2)ysin((pi*x)/2) at the point x=y=1.

for this to find grad f i get 6xy(pi/2)cos((pi*x)/2) i + 3x^2 sin ((pi*x)/2) j

so using the point (1,1,0) this would give grad f as (0,3,0) at this point

so is the equation of the plane (r-(1,1,o))dot(0,3,0)=0 ?
The surface has equation f(x, y, z) = 0, where

f(x, y, z) = 3x^2 y sin((1/2)pi x) - z

Don't forget the "-z". Then

\nabla

f(x, y, z)
= (6xy sin((1/2)pi x) + (3/2)pi x^2 y cos((1/2)pi x), 3x^2 sin((1/2)pi x), -1)

--

When x = y = 1 on the surface, z = 3 (not z = 0 as your answer suggests). The grad of f at that point is

\nabla

f(1, 1, 3) = (6, 3, -1)

Your method for getting an equation of the tangent plane is correct. The tangent plane at (1, 1, 3) has normal vector (6, 3, -1), and hence has equation

(r - (1, 1, 3)).(6, 3, -1) = 0
ie, r.(6, 3, -1) = 6

Jitterbug

ii)Take north to be the direction (0,1,0) and east to be the direction (1,0,0), in which direction will a marble roll if placed on the surface at x=1 y=(1/2)?

assuming these points have to be subs into grad f i think it gives me (0,3,0) so does that mean the marble will roll in the y direction i.e. north?
Let g(x, y) = 3x^2 y sin((1/2)pi x).

For any unit vector u, the derivative wrt s of

g((1, 1/2) + s u)

evaluated at s = 0 is u.