what is the shortest distance from the surface xy+15x+z2=209 to the origin ?

Answers

Answer 1

We can determine the shortest distance from the surface xy+15x+z^2=209 to the origin.

Given the equation of the surface is xy + 15x + z^2 = 209.

Let's determine the shortest distance from the surface to the origin.

The shortest distance between the surface and the origin is given by the perpendicular distance, which can be calculated as follows:

Firstly, we need to determine the gradient of the surface, which is the vector normal to the surface.

For this purpose, we need to write the surface equation in the standard form, which is:xy + 15x + z^2 = 209 xy + 15x + (0)z^2 - 209 = 0 (the coefficients of x, y, and z are a, b, and c).

The gradient of the surface is given by the vector: ∇f = (a, b, c) = (y + 15, x, 2z) at the point P(x, y, z), which is a point on the surface.

Here, the normal vector is ∇f = (y + 15, x, 2z).

Now, let's consider a point A on the surface which is closest to the origin.

Let the coordinates of A be (a, b, c).

Therefore, the position vector of A is given by: OA = ai + bj + ck.

The direction of the position vector is in the direction of the normal vector, and therefore: OA is parallel to ∇f.

Thus, we can write: OA = λ∇f = λ(y + 15)i + λxj + 2λzkWhere λ is a scalar.

Since A lies on the surface, we have: a*b + 15a + c^2 = 209.

We also know that OA passes through the origin.

Therefore, the position vector of A is perpendicular to the direction vector OA.

This gives us: OA·OA = 0⟹ (ai + bj + ck)·(λ(y + 15)i + λxj + 2λzk) = 0

Simplifying this equation gives us:aλ(y + 15) + bλx + c(2λz) = 0

Also, we know that OA passes through the origin.

Therefore, the magnitude of OA is equal to the distance of A from the origin.

Hence, we can write: |OA| = √(a^2 + b^2 + c^2)

The value of λ can be obtained from the equation: aλ(y + 15) + bλx + c(2λz) = 0orλ = -2cz / (b + a(y + 15))

Substituting this value of λ in OA, we get: OA = λ(y + 15)i + λxj + 2λzk= -2cz/(b + a(y + 15)) (y + 15)i - 2cz/(b + a(y + 15)) xj - 4cz^2/(b + a(y + 15))k

Substituting this value of λ in |OA|, we get: |OA| = √[(2cz/(b + a(y + 15)))^2 + (2cz/(b + a(y + 15)))^2 + (4cz^2/(b + a(y + 15)))^2] = 2cz√[(y + 15)^2 + x^2 + 4z^2] / |b + a(y + 15)|

The distance of the point A from the origin is |OA|, which is minimized when the denominator is maximized. The denominator is given by |b + a(y + 15)|.

Thus, we have to maximize the denominator with respect to a and b. The condition for maximum value of the denominator is obtained by differentiating the denominator with respect to a and b separately and equating it to zero. The values of a and b obtained from these equations are substituted in the equation a*b + 15a + c^2 = 209 to obtain the coordinates of the point A, which is closest to the origin.

Hence, we can determine the shortest distance from the surface xy+15x+z^2=209 to the origin.

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Answers

Step-by-step explanation:

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The probability of the train not being at the station on arrival would be:

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Find the probability

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[tex]= 16/(16+3.5)[/tex]

[tex]= 16/19.5[/tex]

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[tex]= 0.82[/tex]

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