Question

Find the angle of inclination θ of the tangent plane to the surface at the given point. x^2+y^2=5,(2,1,4)

Answer 1

To find the angle of inclination for the given surface, calculate the gradient at the point, take the dot product with the x-axis unit vector, and use the cosine inverse to find the angle.

Step-by-step explanation:

The task is to find the angle of inclination θ of the tangent plane to the given surface x2+y2=5 at the point (2,1,4). The angle of inclination is the angle between the tangent plane and the positive x-axis, measured counterclockwise.

To find the angle of inclination, we first need the gradient of the surface at the point of interest. The gradient vector (∇f) at a point is perpendicular to the tangent plane at that point. The components of the gradient vector give us the coefficients of the normal to the plane. The equation of the surface provides us f(x, y, z) = x2 + y2 - 5 = 0, which implies ∇f = (2x, 2y, 0). At the point (2,1,4), this becomes (4, 2, 0).

The angle of inclination θ can be found using the dot product between the normal to the plane and the x-axis unit vector (1,0,0). The cosine of the angle θ is given by the dot product of these two vectors divided by the product of their magnitudes. The angle θ itself is then given by θ = cos-1(∇f · i / |∇f|). Substituting the values, we obtain θ = cos-1(4 · 1 / √(42 + 22)). This yields the angle of inclination.

Answer 2

The angle of inclination θ = 45°.

To find the angle of inclination θ of the tangent plane to the surface at a given point, we can use the gradient vector.

The gradient vector of the surface
`\(f(x, y) = x^2 + y^2 - 5\)` is given by:

`\((f_x, f_y)\)`,

where
`\(f_x\)` and
`\(f_y\)` are the partial derivatives of f with respect to x and y.

`\[ f_x` = 2x

`\[ f_y` = 2y

Now, evaluate these partial derivatives at the given point (2, 1):

`\[ f_x` (2, 1) = 2 x 2 = 4

`\[ f_y` (2, 1) = 2 x 1 = 2

So, the gradient vector at the point (2, 1) is (4, 2). This vector is normal to the tangent plane.

Now, the angle of inclination θ between a vector (a, b) and the positive x-axis is given by:

`\[ \theta = \arctan\left((b)/(a)\right) \]`

Given, a = 4 and b = 2:

`\[ \theta = \arctan\left((2)/(4)\right) = \arctan\left((1)/(2)\right) \]`

θ = 45° or
`(\pi)/(4)`

This represents the angle between the tangent plane to the surface and the positive x-axis at the point (2, 1).