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2 years ago

WILL MARK BRAINLIEST. PLEASE ANSWER FAST

Mathematics

1 answer:

Stella [2.4K]2 years ago

6 0

Answer:

can u copy and paste it

Step-by-step explanation:

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Particle P moves along the y-axis so that its position at time t is given by y(t)=4t−23 for all times t. A second particle, part

sergey [27]

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:

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

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

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

$x(k) = \lim_{k \to 0} \frac{\pi\cdot \sin (2\pi-k)}{k}$

$x(k) = -\pi\cdot \lim_{k \to 0} \frac{\sin k}{k}$

$x(k) = -\pi$

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)

Where $\vec v_{P}(t)$ is the vector velocity of particle P.

If we know that $\vec v_{P}(t) = (0, 4)$ , $\vec v_{Q}(t) = \left(\frac{2\pi\cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, 0 \right)$ and $t = \frac{1}{2}$ , then the vector rate of change of the distance between the two particles:

$\dot r_{P/Q}(t) = \left(\frac{2\pi \cdot \cos \pi t - \pi\cdot t \cdot \cos \pi t + \sin \pi t}{(2-t)^{2}}, -4 \right)$

$\dot r_{Q/P}\left(\frac{1}{2} \right) = \left(\frac{2\pi\cdot \cos \frac{\pi}{2}-\frac{\pi}{2}\cdot \cos \frac{\pi}{2} +\sin \frac{\pi}{2}}{\frac{3}{2} ^{2}}, -4 \right)$

$\dot r_{Q/P} \left(\frac{1}{2} \right) = \left(\frac{4}{9}, -4 \right)$

The magnitude of the vector rate of change is determined by Pythagorean theorem:

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

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

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:

Particle $P$ moves along the y-axis so that its position at time $t$ is given by $y(t) = 4\cdot t - 23$ for all times $t$ . A second particle, $Q$ , moves along the x-axis so that its position at time $t$ is given by $x(t) = \frac{\sin \pi t}{2-t}$ for all times $t \ne 2$ .

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

3 0

1 year ago

Find j(3x^4y^7) if j(n)=n^3

qaws [65]

Answer:

Step-by-step explanation:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

3*x-(4*y-7)=0

STEP

Equation of a Straight Line

1.1 Solve 3x-4y+7 = 0

Tiger recognizes that we have here an equation of a straight line. Such an equation is usually written y=mx+b ("y=mx+c" in the UK).

"y=mx+b" is the formula of a straight line drawn on Cartesian coordinate system in which "y" is the vertical axis and "x" the horizontal axis.

In this formula :

y tells us how far up the line goes

x tells us how far along

m is the Slope or Gradient i.e. how steep the line is

b is the Y-intercept i.e. where the line crosses the Y axis

The X and Y intercepts and the Slope are called the line properties. We shall now graph the line 3x-4y+7 = 0 and calculate its properties

Graph of a Straight Line :

Calculate the Y-Intercept :

Notice that when x = 0 the value of y is 7/4 so this line "cuts" the y axis at y= 1.75000

y-intercept = 7/4 = 1.75000

Calculate the X-Intercept :

When y = 0 the value of x is -7/3 Our line therefore "cuts" the x axis at x=-2.33333

x-intercept = -7/3 = -2.33333

Calculate the Slope :

Slope is defined as the change in y divided by the change in x. We note that for x=0, the value of y is 1.750 and for x=2.000, the value of y is 3.250. So, for a change of 2.000 in x (The change in x is sometimes referred to as "RUN") we get a change of 3.250 - 1.750 = 1.500 in y. (The change in y is sometimes referred to as "RISE" and the Slope is m = RISE / RUN)

Slope = 1.500/2.000 = 0.750

Geometric figure: Straight Line

Slope = 1.500/2.000 = 0.750

x-intercept = -7/3 = -2.33333

y-intercept = 7/4 = 1.75000

4 0

2 years ago

Can some one help me with this question its hard :( down below ill give brainliest

Elanso [62]

Should be D. (5^2)^2= 5^4 (you multiply them) and 5^4+5^4=5^8

4 0

2 years ago

Ronald’s insurance company pays for 80% of his shoulder surgery, after he pays a $500 deductible.

Ivahew [28]

1st a
2nd c
3rd c vchbdnhgd

7 0

3 years ago

I need to solve this can anyone solve this?

xenn [34]

Answer:

d = 7

Step-by-step explanation:

7 times 5 = 35

35/5 = 7

4 0

2 years ago