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NeTakaya
1 year ago
15

7-18 use part 1 of the fundamental theorem of calculus to find the derivative of the function.

Mathematics
1 answer:
blondinia [14]1 year ago
7 0

\displaystyle h(x)=\int\limits_{1}^{\sqrt{x}}~\cfrac{z^2}{z^4+1}dz~\hspace{10em}\cfrac{dh}{du}\cdot \stackrel{chain~rule}{\cfrac{du}{dx}\implies \cfrac{dh}{dx}} \\\\[-0.35em] ~\dotfill\\\\ u=\sqrt{x}\implies \cfrac{du}{dx}=\cfrac{1}{2\sqrt{x}} \\\\[-0.35em] ~\dotfill

\cfrac{dh}{dx}\implies \displaystyle \cfrac{d}{du}\left[ \int\limits_{1}^{u}~\cfrac{z^2}{z^4+1}dz \right]\cdot \cfrac{1}{2\sqrt{x}}\implies \left[ \cfrac{u^2}{u^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}} \\\\\\ \stackrel{substituting~back}{\left[ \cfrac{(\sqrt{x})^2}{(\sqrt{x})^4+1} \right]\cdot \cfrac{1}{2\sqrt{x}}}\implies \cfrac{x}{x^2+1}\cdot \cfrac{1}{2\sqrt{x}}\implies \cfrac{\sqrt{x}}{2x^2+2}

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Step-by-step explanation:

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The point Y is 3 units below the x-axis, so its y-coordinate is -3. This is the value you can read on the y-axis next to point Y.

  (x, y) = (0, -3)

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It is a good idea to learn to read coordinates from a graph. You will be doing it a lot. The x-coordinate is the number of units right of the y-axis. The y-coordinate is the number of units up from the x-axis. Left or down makes the coordinate negative.

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I need help in learning how to solve this equation using substitution:
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2 years ago
The plane x+y+2z=8 intersects the paraboloid z=x2+y2 in an ellipse. Find the points on this ellipse that are nearest to and fart
DiKsa [7]

Answer:

The minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

Step-by-step explanation:

Here, the two constraints are

g (x, y, z) = x + y + 2z − 8  

and  

h (x, y, z) = x ² + y² − z.

Any critical  point that we find during the Lagrange multiplier process will satisfy both of these constraints, so we  actually don’t need to find an explicit equation for the ellipse that is their intersection.

Suppose that (x, y, z) is any point that satisfies both of the constraints (and hence is on the ellipse.)

Then the distance from (x, y, z) to the origin is given by

√((x − 0)² + (y − 0)² + (z − 0)² ).

This expression (and its partial derivatives) would be cumbersome to work with, so we will find the the extrema  of the square of the distance. Thus, our objective function is

f(x, y, z) = x ² + y ² + z ²

and

∇f = (2x, 2y, 2z )

λ∇g = (λ, λ, 2λ)

µ∇h = (2µx, 2µy, −µ)

Thus the system we need to solve for (x, y, z) is

                           2x = λ + 2µx                         (1)

                           2y = λ + 2µy                       (2)

                           2z = 2λ − µ                          (3)

                           x + y + 2z = 8                      (4)

                           x ² + y ² − z = 0                     (5)

Subtracting (2) from (1) and factoring gives

                     2 (x − y) = 2µ (x − y)

so µ = 1  whenever x ≠ y. Substituting µ = 1 into (1) gives us λ = 0 and substituting µ = 1 and λ = 0  into (3) gives us  2z = −1  and thus z = − 1 /2 . Subtituting z = − 1 /2  into (4) and (5) gives us

                            x + y − 9 = 0

                         x ² + y ² +  1 /2  = 0

however, x ² + y ² +  1 /2  = 0  has no solution. Thus we must have x = y.

Since we now know x = y, (4) and (5) become

2x + 2z = 8

2x  ² − z = 0

so

z = 4 − x

z = 2x²

Combining these together gives us  2x²  = 4 − x , so

2x²  + x − 4 = 0 which has solutions

x =  (-1+√33)/4

and

x = -(1+√33)/4.

Further substitution yeilds the critical points  

((-1+√33)/4; (-1+√33)/4; (17-√33)/4)   and

(-(1+√33)/4; - (1+√33)/4; (17+√33)/4).

Substituting these into our  objective function gives us

f((-1+√33)/4; (-1+√33)/4; (17-√33)/4) = (195-19√33)/8

f(-(1+√33)/4; - (1+√33)/4; (17+√33)/4) = (195+19√33)/8

Thus minimum distance of   √((195-19√33)/8)  occurs at  ((-1+√33)/4; (-1+√33)/4; (17-√33)/4)  and the maximum distance of  √((195+19√33)/8)  occurs at (-(1+√33)/4; - (1+√33)/4; (17+√33)/4)

4 0
3 years ago
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