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4 years ago

5

(1 point) Consider the paraboloid z=x2+y2. The plane 2x−2y+z−8=0 cuts the paraboloid, its intersection being a curve. Find "the

natural" parametrization of this curve. Hint: The curve which is cut lies above a circle in the xy-plane which you should parametrize as a function of the variable t so that the circle is traversed counterclockwise exactly once as t goes from 0 to 2*pi, and the paramterization starts at the point on the circle with largest x coordinate. Using that as your starting point, give the parametrization of the curve on the surface.

1 answer:

skad [1K]4 years ago

8 0

Answer:

$\left \{ {{x(t)=-1+\sqrt{10}Sin(t) } \atop {y(t)=1+\sqrt{10}Sin(t)}} \right.$

Step-by-step explanation:

We have the plane 2x-2y+z-8=0 and the paraboloid z=x²+y².

We match the equations:

z=8-2x+2y

z=x²+y²

x²+y²=8-2x+2y

Then, we get the equation of the curve:

x²+2x+y²-2y=8 (adding +2x and -2y in both members of the equation)

(x²+2x+1)+(y²-2y+1)=8+1+1

Trinomials can be reduced as squared binomials

(x+1)²+(y-1)²=10

This is a circle with center in (-1,1) and radius √10

The generic

The parametric expression of the circle is:

$\left \{ {{x(t)=-1+\sqrt{10}Sin(t) } \atop {y(t)=1+\sqrt{10}Sin(t)}} \right.$

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A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the cha

Galina-37 [17]

Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
the radius $r=5cm$
Punch is being poured into the bowl
The height at which the punch is increasing in the bowl is $\frac{dh}{dt} = 1.5$
the exposed area is a circle, (since the bowl is a hemisphere)
the radius of this circle can be written as $'a'$

what is being asked is the rate of change of the exposed area when the height $h = 2 cm$
the rate of change of exposed area can be written as $\frac{dA}{dt}$ .
since the exposed area is changing with respect to the height of punch. We can use the chain rule: $\frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}$

and since $A = \pi a^2$ the chain rule above can simplified to $\frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt}$ -- we can call this Eq(1)

Solution:

the area of the exposed circle is

$A =\pi a^2$

the rate of change of this area can be, (using chain rule)

$\frac{dA}{dt} = 2 \pi a \frac{da}{dt}$ we can call this Eq(2)

what we are really concerned about is how $a$ changes as the punch is being poured into the bowl i.e $\frac{da}{dh}$

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

$r = \frac{a^2 + h^2}{2h}$

and rearrage the formula so that a is the subject:

$a^2 = 2rh - h^2$

now we can derivate a with respect to h to get $\frac{da}{dh}$

$2a \frac{da}{dh} = 2r - 2h$

simplify

$\frac{da}{dh} = \frac{r-h}{a}$

we can put this in Eq(1) in place of $\frac{da}{dh}$

$\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}$

and since we know $\frac{dh}{dt} = 1.5$

$\frac{da}{dt} = \frac{(r-h)(1.5)}{a}$

and now we use substitute this $\frac{da}{dt}$ . in Eq(2)

$\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}$

simplify,

$\frac{dA}{dt} = 3 \pi (r-h)$

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

$r = 5cm$

$h = 2cm$ from the question

$\frac{dA}{dt} = 3 \pi (5-2)$

$\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s$

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vfiekz [6]

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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