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tatyana61 [14]
3 years ago
8

Find the orthogonal trajectories of the family of curves. (Use C for any needed constant.)

Mathematics
1 answer:
Dominik [7]3 years ago
8 0

Answer:

y = Ax^2

Step-by-step explanation:

below is the detailed solution

To determine the orthogonal trajectories of the family of curves

X^2 + 2y^2 = 17k^2

we have to differentiateX^2 + 2y^2 = 17k^2  with respect to x

= 2x + 4y dy/dx = 0

Hence : dy/dx = - x/2y

we have to determine the negative reciprocal

dy/dx = 2y/x ----------- 1

integrate equation 1

∫dy/2y = ∫dx/x

= 1/2 log y = log x + log c

log y = 2logx + 2logc

log y = logx^2 + logC^2

therefore : y = Ax^2  ;   where C^2 = A

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Given the set points K+1:(-1,8),(1,0) and (2,5), find the quadratic polynomial interpolate
Sauron [17]

Answer:

The interpolating polynomial is p(x) = 1-4x+3x^2.

Step-by-step explanation:

We want to find a quadratic polynomial p(x) such that p(-1)=8, p(1)=0 and p(2)=5. In order to do this let us write p(x) = a_0+a_1x+a_3x^2.

Now, evaluating the polynomial in the points -1, 1 and 2 we get

\begin{cases} 8 = p(-1) &= a_0-a_1+a_2\\ 0 = p(1) &= a_0+a_1+a_2\\ 5 = p(2) &= a_0+2a_1+4a_2\end{cases}

This relations give us a linear system of equations:

\begin{cases} 8 &= a_0-a_1+a_2\\ 0 &= a_0+a_1+a_2\\ 5&= a_0+2a_1+4a_2\end{cases}

where the a_0, a_1 and a_2 are the unknowns.

The augmented matrix of the system is

\begin{pmatrix}1 & -1 & 1 & 8\\ 1 & 1 & 1 & 0\\ 1 & 2 & 4 & 5\end{pmatrix}

In this matrix it is easy to eliminate the 1's of the first column and get

\begin{pmatrix} 1 & -1 & 1 & 8\\ 0 & 2 & 0 & -8\\ 0 & 3 & 3 & -3\end{pmatrix}

From this matrix we can find the values of each unknown. Notice that the second row gives us 2a_2=-8 that yields a_1=-4.

Then, the third row means 3a_1+3a_2=-3 that gives -12+3a_2=-3. So, a_2=3.

Finally, the first row is a_0-a_1+a_2=8 and substituting is a_0+7=8 that yields a_0=1.

Therefore, the interpolating polynomial is

p(x) = 1-4x+3x^2.

8 0
3 years ago
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