Mathematics, 24.02.2020 19:05 angelicar4144

The plane x + y + 2z = 30 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on the ellipse that are nearest to and farthest from the origin. g

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Answer from: aubrey1161

z=x^2+y^2

$x+y+2z=30\implies z=\dfrac{30-x-y}2$

$\implies x^2+y^2=\dfrac{30-x-y}2\implies x^2+\dfrac x2+y^2+\dfrac y2=15$

Complete the square to get the equation of the ellipse,

$x^2+\dfrac x2+\dfrac1{16}+y^2+\dfrac y2+\dfrac1{16}=15$

$\implies\left(x+\dfrac14\right)^2+\left(y+\dfrac14\right)^2=\dfrac{121}8$

Parameterize this intersection by the vector-valued function,

$\vec r(t)=\langle x(t),y(t),z(t)\rangle$

$\vec r(t)=\left\langle\sqrt{\dfrac{121}8}\cos t-\dfrac14,\sqrt{\dfrac{121}8}\sin t-\dfrac14,\dfrac{30-\left(\sqrt{\frac{121}8}\cos t-\frac14\right)-\left(\sqrt{\frac{121}8}\sin t-\frac14\right)}2\right\rangle$

$\vec r(t)=\left\langle\sqrt{\dfrac{121}8}\cos t-\dfrac14,\sqrt{\dfrac{121}8}\sin t-\dfrac14,\dfrac{241}{16}-\sqrt{\dfrac{121}{32}}(\cos t+\sin t)\right\rangle$

where $0\le t\le2\pi$ .

The distance between the origin and any point on the curve is $\|\vec r(t)\|$ , which is

$\|\vec r(t)\|=\sqrt{x(t)^2+y(t)^2+z(t)^2}$

Recall that a function f(x) and its square root $\sqrt{f(x)}$ share the same critical points, so we can make things easier by looking for the critical points of $\|\vec r(t)\|^2$ . Let

$f(t)=\|\vec r(t)\|^2$

$f(t)=\left(\sqrt{\dfrac{121}8}\cos t-\dfrac14\right)^2+\left(\sqrt{\dfrac{121}8}\sin t-\dfrac14\right)^2+\left(\dfrac{241}{16}-\sqrt{\dfrac{121}{32}}(\cos t+\sin t)\right)^2$

$f(t)=\dfrac{8051}{32}-\dfrac{693}{8\sqrt2}(\cos t+\sin t)+\dfrac{121}{16}\sin t\cos t$

Differentiate and find the critical points:

$f'(t)=\dfrac{693}{8\sqrt2}(\sin t-\cos t)+\dfrac{121}{16}(\cos^2t-\sin^2t)$

$f'(t)=\dfrac1{16\sqrt2}(\sin t-\cos t)(1386+121\sqrt2(\cos t+\sin t))=0$

$\implies\sin t=\cos t\implies t=\dfrac\pi4,\dfrac{5\pi}4$

The distance from the origin to the critical points of $\vec r(t)$ are then

$\left\|\vec r\left(\dfrac\pi4\right)\right\|=\dfrac{15\sqrt3}2$

$\left\|\vec r\left(\dfrac{5\pi}4\right)\right\|=3\sqrt{38}$

Then the point closest to the origin is

$\left(x\left(\dfrac\pi4\right),y\left(\dfrac\pi4\right),z\left(\dfrac\pi4\right)\right)=\left(\dfrac52,\dfrac52,\dfrac{25}2\right)$

and the point furthest from the origin is

$\left(x\left(\dfrac{5\pi}4\right),y\left(\dfrac{5\pi}4\right),z\left(\dfrac{5\pi}4\right)\right)=(-3,-3,18)$

Answer from: Quest

figure c

step-by-step explanation:

hope this .

Wich of these figures represents a line ?

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

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The plane x + y + 2z = 30 intersects the paraboloid z = x2 + y2 in an ellipse. Find the points on th...

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