calc_tutorial_16_4_010

 
Problem Statement

{3,5,7,9}
pow(7,2)
49-1
2*7

Calculate Tr, Tθ, and n(r, θ) for the parametrized surface at the given point. Then find the equation of the tangent plane to the surface at that point.

G(r, θ) = (r cos θ, r sin θ, 1 - r2); r = 1 / 7, θ = π / 4

 
Step 1

Question Sequence

Question 1

Recall how to find the tangent vectors, Tr, Tθ, and the normal vector, n(r, θ), to a given parametrized surface, G(r, θ).

Tr =

A.
B.
C.

Tθ =

A.
B.
C.

n(r, θ) = Tr × Tθ

Correct.
Incorrect.

 
Step 2

Question Sequence

Question 2

Find the general expressions for Tr and Tθ given G(r, θ) = (r cos θ, r sin θ, 1 - r2)

Tr = (∂ / ∂r)(r cos θ, r sin θ, 1 - r2)

= <cos θ, sin θ, r>

Tθ = (∂ / ∂θ)(r cos θ, r sin θ, 1 - r2)

= <-r sin θ, r cos θ, >

Calculate Tr and Tθ at the point where r = 1 / 7 and θ = π / 4

Tr = <1/√2, 1/√2, -2/>

Tθ = <-1/(√2), 1/(√2), 0>

Correct.
Incorrect.

 
Step 3

Question Sequence

Question 3

Find the general expression for n(r, θ). Let the table below represent the matrix A

i j k
cos θ sin θ -2r
-r sin θ r cos θ 0
Table : Matrix A

n(r, θ) = Tr × Tθ

= <cos θ, sin θ, -2r> × <-r sin θ, r cos θ, 0>

= det(A)

= r<r cos θ,r sin θ, >

Calculate n(r, θ) at the point when r = 1 / 7 and θ = π / 4

n(1 / 7, π / 4) = <√2/, √2/, 1/>

Correct.
Incorrect.

 
Step 4

Question Sequence

Question 4

We now wish to find the equation of the tangent plane to the surface G(r, θ) = (r cos θ, r sin θ, 1 - r2) at the point where r = 1 / 7 and θ = π / 4.

Recall that an equation of the plane through the point P = <x0, y0, z0> with normal vector n can be described as follows.

<x - x0, y - y0, z - z0> · n = 0

Find the coordinates of the point on the surface G(r, θ) at the point where r = 1 / 7 and θ = π / 4.

P(x, y, z) = ((1/7)cos(π/4), (1/7)sin(π/4), 1 - (1/7)2)

= (1/(√2), 1/(√2), /49)

Correct.
Incorrect.

 
Step 5

Question Sequence

Question 5

Find the equation of the tangent plane to G(r, θ) = (r cos θ, r sin θ, 1 - r2) at the point where r = 1 / 7 and θ = π / 4

We will use the equation <x - x0, y - y0, z - z0> · n = 0. The point on our surface is P = (1/(7√2), 1/(7√2), 48/49) and the normal vector to the surface at that point is n = (1/7)<√2/7, √2/7, 1>. It suffices to use the normal vector n = <√2/7, √2/7, 1> since the scalar factor of 1/7 will not affect the equation of the plane.

Write the equation of the tangent plane by computing the dot product.

<x - x0, y - y0, z - z0> · n = 0

<x - (1/(7√2), y - 1/(7√2), z - (48/49)> · <√2/7, √2/7, 1> = 0

(√2/)(x - (√2/)) + (√2/7)(y - (√2/)) + z - (48/) = 0

Correct.
Incorrect.