Find parametric equations of the curve given by the intersection of the surfaces:

The cone: z=sqrt(x^2+y^2)


The plane: z=4+y

x(t)=
y(t)=
z(t)=

answer is not sqrt(16+8t);t;4+t
.

To find the parametric equations for the curve of intersection between the surfaces:

1. **Cone equation:** \( z = \sqrt{x^2 + y^2} \)
2. **Plane equation:** \( z = 4 + y \)

### Step 1: Set the \( z \)-coordinates equal
Since both surfaces share the same \( z \)-value at their intersection, we can set the equations for \( z \) equal to each other:

\[
\sqrt{x^2 + y^2} = 4 + y
\]

### Step 2: Square both sides
To eliminate the square root, square both sides of the equation:

\[
x^2 + y^2 = (4 + y)^2
\]

Expanding the right-hand side:

\[
x^2 + y^2 = 16 + 8y + y^2
\]

### Step 3: Simplify the equation
Subtract \( y^2 \) from both sides to eliminate \( y^2 \):

\[
x^2 = 16 + 8y
\]

This simplifies to:

\[
x^2 = 8(y + 2)
\]

### Step 4: Solve for \( x \)
We can now solve for \( x \):

\[
x = \pm \sqrt{8(y + 2)}
\]

For simplicity, choose the positive root:

\[
x = \sqrt{8(y + 2)}
\]

### Step 5: Parametrize \( y \)
We can let \( y \) be a parameter. Let's define \( y = t \), so the parametric equations will be in terms of \( t \). Thus:

\[
x = \sqrt{8(t + 2)}
\]
\[
y = t
\]

### Step 6: Parametrize \( z \)
Now, using the plane equation \( z = 4 + y \), we substitute \( y = t \):

\[
z = 4 + t
\]

### Final Parametric Equations
The parametric equations for the curve of intersection are:

\[
x(t) = \sqrt{8(t + 2)}
\]
\[
y(t) = t
\]
\[
z(t) = 4 + t
\]

Let me know if you would like further details or clarifications! 

Here are 5 related questions to further explore the topic:
1. How do the parametric equations describe the curve geometrically?
2. What happens to the curve if the plane equation is modified to \( z = 5 + y \)?
3. Can you parametrize the curve using polar coordinates?
4. How would the curve change if the cone equation was \( z = 2\sqrt{x^2 + y^2} \)?
5. What is the geometric significance of the parameter \( t \) in this case?

**Tip:** When squaring both sides of an equation, always check for extraneous solutions introduced by the squaring process!

Find parametric equations of the curve given by the intersection of the surfaces:

The cone: z=sqrt(x^2+y^2)

The plane: z=4+y

x(t)=
y(t)=
z(t)=

answer is not sqrt(16+8t);t;4+t.

Explore the parametric equations for the curve formed by the intersection of the cone z = sqrt(x^2 + y^2) and the plane z = 4 + y. This problem involves parametric equations, conic sections, and surface intersection, suitable for undergraduate-level calculus students.

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