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3 years ago

Ava wants to draw a parallelogram on the coordinate plane. She

Mathematics

1 answer:

wolverine [178]3 years ago

8 0

Answer:

$k =(2,1)$

$JK = 2$

Step-by-step explanation:

Given

$J = (0,1)$ ---- $(x_1,y_1)$

$H = (1,-2)$ --- $(x_2,y_2)$

$I = (3,-2)$ --- $(x_3,y_3)$

See attachment for grid

Solving (a): The coordinates of K

The parallelogram has the following diagonals: IJ and HK

Diagonals bisect one another. So:

Midpoint of IJ = Midpoint of HK

This gives:

$\frac{1}{2}(I + J) = \frac{1}{2}(H+K)$

$\frac{1}{2}(x_3+x_1,y_3+y_1) = \frac{1}{2}(x_2+x,y_2+y)$

$\frac{1}{2}(3+0,-2+1) = \frac{1}{2}(1+x,-2+y)$

$\frac{1}{2}(3,-1) = \frac{1}{2}(1+x,-2+y)$

Multiply through by 2

$(3,-1) = (1+x,-2+y)$

By comparison:

$1 + x = 3$

$-2 + y = -1$

Solve for x and y

$x = 3 - 1 =2$

$y = -1 +2 = 1$

So, the coordinates of k is:

$k =(2,1)$

The length of JK is calculated using distance (d) formula

$d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$

$J = (0,1)$ ---- $(x_1,y_1)$

$k =(2,1)$ ---- $(x_2,y_2)$

So:

$d = \sqrt{(0 - 2)^2 + (1 - 1)^2$

$d = \sqrt{(- 2)^2 + (0)^2$

$d = \sqrt{4 + 0$

$d = \sqrt{4$

$d = 2$

Hence:

$JK = 2$

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a) The limit of the position of particle $Q$ when time approaches 2 is $-\pi$ .

b) The velocity of particle $Q$ is $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$ for all $t \ne 2$ .

c) The rate of change of the distance between particle $P$ and particle $Q$ at time $t = \frac{1}{2}$ is $\frac{4\sqrt{82}}{9}$ .

<h3>How to apply limits and derivatives to the study of particle motion</h3>

a) To determine the limit for $t = 2$ , we need to apply the following two algebraic substitutions:

$u = \pi t$ (1)

$k = 2\pi - u$ (2)

Then, the limit is written as follows:

$x(t) = \lim_{t \to 2} \frac{\sin \pi t}{2-t}$

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$x(k) = -\pi$

The limit of the position of particle $Q$ when time approaches 2 is $-\pi$ . $\blacksquare$

b) The function velocity of particle $Q$ is determined by the derivative formula for the division between two functions, that is:

$v_{Q}(t) = \frac{f'(t)\cdot g(t)-f(t)\cdot g'(t)}{g(t)^{2}}$ (3)

Where:

$f(t)$ - Function numerator.
$g(t)$ - Function denominator.
$f'(t)$ - First derivative of the function numerator.
$g'(x)$ - First derivative of the function denominator.

If we know that $f(t) = \sin \pi t$ , $g(t) = 2 - t$ , $f'(t) = \pi \cdot \cos \pi t$ and $g'(x) = -1$ , then the function velocity of the particle is:

$v_{Q}(t) = \frac{\pi \cdot \cos \pi t \cdot (2-t)-\sin \pi t}{(2-t)^{2}}$

$v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$

The velocity of particle $Q$ is $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t -\sin \pi t}{(2-t)^{2}}$ for all $t \ne 2$ . $\blacksquare$

c) The vector rate of change of the distance between particle P and particle Q ( $\dot r_{Q/P} (t)$ ) is equal to the vectorial difference between respective vectors velocity:

$\dot r_{Q/P}(t) = \vec v_{Q}(t) - \vec v_{P}(t)$ (4)

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$|\dot r_{Q/P}| = \sqrt{\left(\frac{4}{9} \right)^{2}+(-4)^{2}}$

$|\dot r_{Q/P}| = \frac{4\sqrt{82}}{9}$

The rate of change of the distance between particle $P$ and particle $Q$ at time $t = \frac{1}{2}$ is $\frac{4\sqrt{82}}{9}$ . $\blacksquare$

<h3>Remark</h3>

The statement is incomplete and poorly formatted. Correct form is shown below:

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a) As times approaches 2, what is the limit of the position of particle $Q?$ Show the work that leads to your answer.

b) Show that the velocity of particle $Q$ is given by $v_{Q}(t) = \frac{2\pi\cdot \cos \pi t-\pi\cdot t \cdot \cos \pi t +\sin \pi t}{(2-t)^{2}}$ .

c) Find the rate of change of the distance between particle $P$ and particle $Q$ at time $t = \frac{1}{2}$ . Show the work that leads to your answer.

To learn more on derivatives, we kindly invite to check this verified question: brainly.com/question/2788760

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2 years ago