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Mathematics

1 answer:

pashok25 [27]3 years ago

5 0

Here's li $^{}$ nk to the answer:

cutt $^{}$ .ly/4Rq $^{}$ tIvk

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At what point on the paraboloid y = x2 + z2 is the tangent plane parallel to the plane 3x + 2y + 7z = 2? (if an answer does not

Nikitich [7]

If $f(x, y, z) = c$ represent a family of surfaces for different values of the constant $c$ . The gradient of the function $f$ defined as $\nabla f$ is a vector normal to the surface $f(x, y, z) = c$ .

Given the paraboloid

$y = x^2 + z^2$ .

We can rewrite it as a scalar value function f as follows:

$f(x,y,z)=x^2-y+z^2=0$

The normal to the paraboloid at any point is given by:

$\nabla f= i\frac{\partial}{\partial x}(x^2-y+z^2) - j\frac{\partial}{\partial y}(x^2-y+z^2) + k\frac{\partial}{\partial z}(x^2-y+z^2) \\ \\ =2xi-j+2zk$

Also, the normal to the given plane $3x + 2y + 7z = 2$ is given by:

$3i+2j+7k$

Equating the two normal vectors, we have:

$2x=3\Rightarrow x= \frac{3}{2} \\ \\ -1=2 \\ \\ 2z=7\Rightarrow z= \frac{7}{2}$

Since, -1 = 2 is not possible, therefore there exist no such point on the paraboloid $y = x^2 + z^2$ such that the tangent plane is parallel to the plane 3x + 2y + 7z = 2.

4 0

4 years ago

Find the 50th term of the sequence 5, -2, -9, -16

VARVARA [1.3K]

Answer:

-35

Step-by-step explanation:

it is going backwards by -7 so 5x-7 = -35

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3 years ago

Given: ABCD is a trapezoid, AC ⊥ CD AB = CD, AC=the square root of 75 , AB = 5 Find: AABCD

Ugo [173]

In this attached picture according to the conditions of the problem we have an isosceles trapezoid and since we know that legs are equal (AD=BC=5 cm), we have to calculate bases and height in order to find the area. Working with the triangle BCD, we apply Pythagoras theorem and find that CD = $\sqrt{75+25}$ = 10 cm. Since BDC is a right triangle, applying theorem for the area of triangles, we find that $\frac{1}{2} * BF = \frac{1}{2} * 5 * \sqrt{75}$ and BF= $0.5\sqrt{75}$ . Since ABCD is an isosceles trapezoid, triangles ADE and BFC are congruent with Angle Side Angle theorem. Then, DE=FC and with the help of Pythagoras theorem, DE=FC=2.5 cm. Then, AB=EF=5 cm and the area of the trapezoid is $A= BF * \frac{AB+CD}{2} = 0.5 \sqrt{75} * \frac{5+10}{2} = 18.75 \sqrt{3} cm^{2}$