Find the surface area of that part of the plane 9x 5y z=5 that lies inside the elliptic cylinder x216 y281=1

1 answer:

3 0

I assume the equation of the plane is $9x+5y+z=5$ and the cylinder has equation $\dfrac{x^2}{16}+\dfrac{y^2}{81}=1$ .

I don't know what techniques are available to you, so I'll resort to (in my opinion) the most reliable: surface integration.

The surface of intersection $S$ is an ellipse in three dimensional space which can be parameterized by $\mathbf x(u,v)=\langle 4u\cos v,9u\sin v,5-36u\cos v-45 u\sin v\rangle$ , with $u\in[0,1]$ and $v\in[0,2\pi]$ .

The area is then given by the integral
$\displaystyle\iint_S\mathrm dA=\int_0^1\int_0^{2\pi}\left\|\dfrac{\partial\mathbf x}{\partial u}\times\dfrac{\partial\mathbf x}{\partial v}\right\|\,\mathrm dv\,\mathrm du=36\sqrt{107}\pi$

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How do you do this question?

vaieri [72.5K]

Answer:

(-∞, -2), (-2, -0.618), and (1.618, 3)

Step-by-step explanation:

The red plus (+) signs indicate the regions in which the function is concave up, and the red negative (-) signs indicate the regions in which the function is concave down.

Note that the sign of the concavity changes at an inflection point.

Let's examine the intervals given.

(-∞, -2): Yes, concave up.

(-∞, -1.17): No. Concave up in (-∞, -2) but concave down in (-2, -1.17).

(-2, 0): No. Concave down in (-2, -1.17) but increasing in (-1.17, 0.0).

(-1.17, 0.689): Yes. Concave up.

(-0.618, 1.618): No. Concave up in (-0.618, 0.689) but concave down in (0.689, 1.618).

(0, 3): No. Concave up in (0, 0.689), concave down in (0.689, 2.481), and concave up in (2.481, 3).

(2.481, ∞): Yes. Concave up.

The three intervals that are concave up are (-∞, -2), (-1.17, -0.689), and (2.481, ∞).

4 0

3 years ago

Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer

agasfer [191]

Answer:

No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming: the function is $f(x)=x^{2}$ in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

$g'(0)=g'(1)$

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

$g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0$

$g(x)=x^{2}\\g'(x)=2x\\ g'(1)=2(1) \Rightarrow g'(1)=2$

g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

3) Since g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Because this is the same as to calculate the limit from the left and right side, of g(x).

$f'(c)=\lim_{x\rightarrow c}\left [\frac{f(b)-f(a)}{b-a} \right ]\\\\g'(0)=\lim_{x\rightarrow 0}\left [\frac{g(b)-g(a)}{b-a} \right ]\\\\g'(1)=\lim_{x\rightarrow 1}\left [\frac{g(b)-g(a)}{b-a} \right ]$