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2 years ago

Find three different surfaces that contain the curve r(t) = t^2 i + lnt j + (1/t)k

Mathematics

1 answer:

Viefleur [7K]2 years ago

7 0

solution:

Consider the curve: r(t) = t²i +(int)j + 1/t k

X= t² , y = int ,z = 1/t

Using, x = t², z = 1/t

X = (1/z)²

Xz²= 1

Using y = int, z= 1/t

Y = in│1/z│

Using x = t², y = int

Y = int

= in(√x)

Hence , the required surface are,

Xz² = 1

Y = in│1/z│

Y= in(√x)

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22.Lenin is preparing dinner plates. He has 12 pieces of chicken and 16 rolls. If he wants to make all the plates identicalwitho

Nitella [24]

Answer: The greatest number of plates Lenin can prepare = 4

and there will be 3 chickens and 4 rolls in each plate.

Step-by-step explanation:

Given: Lenin is preparing dinner plates. He has 12 pieces of chicken and 16 rolls.

To make all the plates identical without any food left over, the greatest number of plates Lenin can prepare = G.C.D.(12,16)=4

The number of pieces of chicken in each plate = $\dfrac{12}{4}=3$

The number of pieces of rolls in each plate = $\dfrac{16}{4}=4$

So, the greatest number of plates Lenin can prepare = 4

and there will be 3 chickens and 4 rolls in each plate.

7 0

1 year ago

If the length of rectangle is 8.26cm and its breadth is 5.5cm, the find the

zubka84 [21]

Answer:

Area of a rectangle= L×B

=8.26cm×5.5cm

=45.43cm square

3 0

2 years ago

Read 2 more answers

The extract of a plant native to Taiwan has been tested as a possible treatment for Leukemia. One of the chemical compounds prod

True [87]

Answer:

a) 57.35%

b) 99.99%

c) 68.27%

Step-by-step explanation:

When we have a random variable X that is normally distributed with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ , then

The probability that the random variable has a value less than a, P(X < a) = P(X ≤ a) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ to the left of a.

The probability that the random variable has a value greater than b, P(X > b) = P(X ≥ b) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ to the right of b.

The probability that the random variable has a value between a and b, P(a < X < b) = P(a ≤ X ≤ b) = P(a < X ≤ b)= P(a ≤ X < b) is the area under the normal curve with mean $\large\bf \mu$ and standard deviation $\large\bf \sigma$ between a and b.

In this case, the random variable is the collagen amount found in the extract of the plant. The mean is 63 g/ml and the standard deviation is 5.4 g/ml

(a) What is the probability that the amount of collagen is greater than 62 grams per mililiter?

As we have seen, we need to find the area under the normal curve with mean 63 and standard deviation 5.4 to the right of 62 (see picture).

You can find this value easily with a calculator or a spreadsheet. If you prefer the old-style, then you have to standardize the values and look up in a table.

If you have access to Excel or OpenOffice Calc, you can find this value by introducing the formula:

1- NORMDIST(62,63,5.4,1) in Excel

1 - NORMDIST(62;63;5.4;1) in OpenOffice Calc

and we will get a value of 0.5735 or 57.35%

(b) What is the probability that the amount of collagen is less than 90 grams per mililiter?

Now we want the area to the left of 90

NORMDIST(90,63,5.4,1) in Excel

NORMDIST(90;63;5.4;1) in OpenOffice Calc

You will get a value of 0.9999 or 99.99%

(c) What percentage of compounds formed from the extract of this plant fall within 1 standard deviations of the mean?

You can use either the rule that 68.27% of the data falls between $\large\bf \mu -\sigma$ and $\large\bf \mu +\sigma$ or compute area between 63 - 5.4 and 63 + 5.4, that is to say, the area between 57.6 and 68.4

In Excel

NORMDIST(68.4,63,5.4,1) - NORMDIST(57.6,63,5.4,1)

In OpenOffice Calc

NORMDIST(68.4;63;5.4;1) - NORMDIST(57.6;63;5.4;1)

In any case we get a value of 0.6827 or 68.27%

3 0

2 years ago

Which statements are true Select three options.

BARSIC [14]

The line x = 0 is perpendicular to the line y = -3:

Correct. Any horizontal line (y = a) and any vertical line (x = b) intersect at some point and are perpendicular.

All lines that are parallel to the y-axis are vertical lines:

Correct. The y-axis is a vertical line, so any lines that are parallel to it must also be vertical.

All lines that are perpendicular to the x-axis have a slope of 0.

Incorrect. Lines that have a slope of 0 are horizontal, and the x-axis is horizontal as well. Any lines with a slope of 0 are parallel to the x-axis, not perpendicular to it.

The equation of the line parallel to the x-axis that passes through the point (2, 6) is x = 2.

Incorrect. x = 2 is a vertical line, and vertical lines cannot be parallel to the horizontal x-axis. x = 2 is perpendicular to the x-axis, however.

The equation of the line perpendicular to the y-axis that passes through the point (-5, 1) is y = 1.

Correct. The line y = 1 is horizontal, and the y-axis is a vertical line. Because the line y = 1 crosses the y-axis, the lines are perpendicular.

4 0

2 years ago

Power series of y''+x^2y'-xy=0

Ray Of Light [21]

Assuming we're looking for a power series solution centered around $x=0$ , take

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Substituting into the ODE yields

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}+\sum_{n\ge1}na_nx^{n+1}-\sum_{n\ge0}a_nx^{n+1}=0$

The first series starts with a constant term; the second series starts at $x^2$ ; the last starts at $x^1$ . So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a $x^2$ term. We have

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+6a_3x+\sum_{n\ge4}n(n-1)a_nx^{n-2}$

$\displaystyle\sum_{n\ge0}a_nx^{n+1}=a_0x+\sum_{n\ge1}a_nx^{n+1}$

Re-index the first sum to have it start at $n=1$ (to match the the other two sums):

$\displaystyle\sum_{n\ge4}n(n-1)a_nx^{n-2}=\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}$

So now the ODE is

$\displaystyle\left(2a_2+6a_3x+\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n\ge1}na_nx^{n+1}-\left(a_0x+\sum_{n\ge1}a_nx^{n+1}\right)=0$

Consolidate into one series starting $n=1$ :

$\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0$

Suppose we're given initial conditions $y(0)=a_0$ and $y'(0)=a_1$ (which follow from setting $x=0$ in the power series representations for $y$ and $y'$ , respectively). From the above equation it follows that

$\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_{n+3}+(n-1)a_n=0&\text{for }n\ge2\end{cases}$

Let's first consider what happens when $n=3k-2$ , i.e. $n\in\{1,4,7,10,\ldots\}$ . The recurrence relation tells us that

$a_4=-\dfrac{1-1}{(1+3)(1+2)}a_1=0\implies a_7=0\implies a_{10}=0$

and so on, so that $a_{3k-2}=0$ except for when $k=1$ .

Now let's consider $n=3k-1$ , or $n\in\{2,5,8,11,\ldots\}$ . We know that $a_2=0$ , and from the recurrence it follows that $a_{3k-1}=0$ for all $k$ .

Finally, take $n=3k$ , or $n\in\{0,3,6,9,\ldots\}$ . We have a solution for $a_3$ in terms of $a_0$ , so the next few terms ( $k=2,3,4$ ) according to the recurrence would be

$a_6=-\dfrac2{6\cdot5}a_3=-\dfrac2{6\cdot5\cdot3\cdot2}a_0=-\dfrac{a_0}{6\cdot3\cdot5}$
$a_9=-\dfrac5{9\cdot8}a_6=\dfrac{a_0}{9\cdot6\cdot3\cdot8}$
$a_{12}=-\dfrac8{12\cdot11}a_9=-\dfrac{a_0}{12\cdot9\cdot6\cdot3\cdot11}$

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for $n=3k$ as

$a_{3k}=\dfrac{(-1)^{k+1}a_0}{(3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

So the series solution to the ODE is given by

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y=a_1x+\displaystyle\sum_{k\ge0}\frac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at $a_0=y(0)=1$ and $a_1=y'(0)=2$ overlaid with the series solution (orange) with $n=3$ and $n=6$ . (Note the rapid convergence.)

7 0

2 years ago