Find the slope of the tangent line to the curve of intersection of the

Question

Find the slope of the tangent line to the curve of intersection of the

surface 36𝑥² − 9𝑦² + 4𝑧² + 36 = 0
with the plane 𝑥 = 1 at the point (1, √12, −3) .

Answer 1

at x=1, we have

36 - 9y^2 + 4z^2 + 36 = 0
y^2/8 - z^2/18 = 1
If by slope you mean dy/dz, then
y/4 dy/dz - z/9 = 0
√3/2 dy/dz = -1/3
dy/dz = -2/√27

Answer 2

Well, finding the slope of a tangent line sounds like a slippery slope to me! But I'll give it a shot.

To find the slope of the tangent line, we need to find the partial derivatives of the equation with respect to x, y, and z. Then we can use these partial derivatives to find the slope.

Let's start by finding the partial derivative with respect to x:

Partial derivative of 36x^2 - 9y^2 + 4z^2 + 36 = 0 with respect to x:
72x = 0

Since we're given that x = 1 at the point (1, √12, -3), we know that the partial derivative with respect to x is 72(1) = 72.

Now let's find the partial derivative with respect to y:

Partial derivative of 36x^2 - 9y^2 + 4z^2 + 36 = 0 with respect to y:
-18y = 0

Since we're not given the value of y at the point (1, √12, -3), we can't determine the partial derivative with respect to y.

Lastly, let's find the partial derivative with respect to z:

Partial derivative of 36x^2 - 9y^2 + 4z^2 + 36 = 0 with respect to z:
8z = 0

Since we're not given the value of z at the point (1, √12, -3), we can't determine the partial derivative with respect to z either.

So, unfortunately, we can only determine the slope in the x-direction. Therefore, the slope of the tangent line to the curve of intersection of the given surface and the plane x = 1 at the point (1, √12, -3) is 72 in the x-direction.

Hope that answers your question in a curve-acious way!

Answer 3

To find the slope of the tangent line to the curve of intersection, we need to find the gradient vector of the surface and the plane at the given point (1, √12, -3).

First, let's find the gradient vector of the surface. The gradient vector of a surface is given by the vector (∂𝑓/∂𝑥, ∂𝑓/∂𝑦, ∂𝑓/∂𝑧), where 𝑓(𝑥, 𝑦, 𝑧) is the equation of the surface.

Given the surface equation 36𝑥² − 9𝑦² + 4𝑧² + 36 = 0, we can find the partial derivatives as follows:

∂𝑓/∂𝑥 = 72𝑥
∂𝑓/∂𝑦 = -18𝑦
∂𝑓/∂𝑧 = 8𝑧

Plugging in the point (1, √12, -3) into these partial derivatives, we get:

∂𝑓/∂𝑥 = 72(1) = 72
∂𝑓/∂𝑦 = -18√12
∂𝑓/∂𝑧 = 8(-3) = -24

Therefore, the gradient vector of the surface at the point (1, √12, -3) is (72, -18√12, -24).

Next, let's find the gradient vector of the plane 𝑥 = 1. Since the equation of the plane is already given in terms of 𝑥, the gradient vector is (1, 0, 0).

Now, to find the slope of the tangent line to the curve of intersection, we take the dot product of the gradient vectors of the surface and the plane:

Gradient vector of the surface (72, -18√12, -24) dot Gradient vector of the plane (1, 0, 0):

(72)(1) + (-18√12)(0) + (-24)(0) = 72

Therefore, the slope of the tangent line to the curve of intersection of the surface with the plane 𝑥 = 1 at the point (1, √12, −3) is 72.

Answer 4

To find the slope of the tangent line to the curve of intersection of the given surface and the plane, we first need to determine the equation of the curve of intersection.

The equation of the surface is given by:
36𝑥² − 9𝑦² + 4𝑧² + 36 = 0

And the equation of the plane is given by:
𝑥 = 1

To find the curve of intersection, we substitute 𝑥 = 1 into the equation of the surface:
36(1)² − 9𝑦² + 4𝑧² + 36 = 0
36 − 9𝑦² + 4𝑧² + 36 = 0
-9𝑦² + 4𝑧² + 72 = 0

Now we have an equation in terms of 𝑦 and 𝑧. To make it easier to work with, let's rearrange the equation:
4𝑧² − 9𝑦² = -72

Next, we need to find the values of 𝑦 and 𝑧 at the point (1, √12, −3).

Substituting 𝑥 = 1 into the equation of the original surface:
36(1)² − 9𝑦² + 4(−3)² + 36 = 0
36 - 9𝑦² + 36 = 0
-9𝑦² + 72 = 0
-9𝑦² = -72
𝑦² = 8
𝑦 = √8 = 2√2

Substituting 𝑥 = 1 and 𝑦 = 2√2 into the equation of the original surface:
36(1)² − 9(2√2)² + 4𝑧² + 36 = 0
36 - 9(8) + 4𝑧² + 36 = 0
36 - 72 + 4𝑧² + 36 = 0
4𝑧² = 0
𝑧 = 0

So the curve of intersection can be represented by the parametric equations:
𝑥 = 1
𝑦 = 2√2
𝑧 = 0

Now, to find the slope of the tangent line at the point (1, √12, −3), we need to find the derivatives of the parametric equations with respect to the parameter.

The derivative of 𝑦 with respect to the parameter is:
𝑑𝑦/𝑑𝑡 = 0

The derivative of 𝑧 with respect to the parameter is:
𝑑𝑧/𝑑𝑡 = 0

Since 𝑥 is constant, its derivative with respect to the parameter is zero as well:
𝑑𝑥/𝑑𝑡 = 0

Therefore, the tangent line at the point (1, √12, −3) is parallel to the 𝑥-axis, and its slope is zero.