Mathematics: Need some with higher level mathematics integration question as attached in this post, specifically parts (iii) and (iv). you in advance to whoever who manages to ! : )

Need some with higher level mathematics integration

Ответ:

lakenyahar

11.01.2022

i. We were given the parametric equations,

$x=e^{-t}, y=e^{-t}sin(t)$

To find the equation of the normal, we need the gradient of the tangent to the cure at P. Then we take the negative reciprocal of it. So let us differentiate to get,

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{dy}{dx} =\frac{e^{-t}(cos(t)-sin(t)}{-e^{-t}}$

$\frac{dy}{dx}=-(cos(t)-sin(t))$

We now substitute

$t=\frac{\pi}{2}$

$\frac{dy}{dx}=-(cos(\frac{\pi}{2})-sin(\frac{\pi}{2}))$

$\frac{dy}{dx}=-(0-1)$

$\frac{dy}{dx}=-(-1)$

$\frac{dy}{dx}=1$

Hence the slope of the normal,

$m=\frac{-1}{1}=-1$

We need one more thing to find the equation of this straight line. Do you remember what that is?

Good! We need a point, so we substitute,

$t=\frac{\pi}{2}$

into

$x=e^{-t}, y=e^{-t}sin(t)$

to get,

$x=e^{-\frac{\pi}{2}}, y=e^{-\frac{\pi}{2}}sin(\frac{\pi}{2})$

This gives us,

$x=e^{-\frac{\pi}{2}}, y=e^{-\frac{\pi}{2}}$

We now use the formula,

y-y_1=m(x-x_1)

to get the equation of the normal.

We substitute to obtain,

$y-e^{-\frac{\pi}{2}}=-1(x-e^{-\frac{\pi}{2}})$

This simplifies to,

$y=-x+e^{-\frac{\pi}{2}}+e^{-\frac{\pi}{2}}$

$y=-x+2e^{-\frac{\pi}{2}}$

iii) From the diagram(Please check second attachment), the normal is above the curve, C on the interval,

$0\le t\le \pi$

So the area,

$A=\int\limits^\pi_0 {y(t)} \, dt -\int\limits^\pi_0 {y(t)\times x'(t)} \, dt$

Meaning that we have to plug in the parametric forms of the equations.

This implies that

$A=\int\limits^\pi_0{-e^{-t}+2e^{-\frac{\pi}{2}}\,dt -\int\limits^\pi_0 {e^{-t}sin(t)\times -e^{-t}\,dt$

This simplifies to;

$A=\int\limits^\pi_0 {-e^{-t}+2e^{-\frac{\pi}{2}}\, dt -\int\limits^\pi_0 {-e^{-2t}sin(t) \, dt$

Now we apply integration by parts(which was given to you already) to evaluate the first integral and then integrate the second integral directly.

$A=(e^{-t}+2e^{-\frac{\pi}{2}})|_0^{\pi}--\frac{2}{5}e^{-2t}sin(t)-\frac{1}{5}e^{-2t} |_0^{\pi}$

Evaluating this gives us,

$A=2\pi e^{-\frac{\pi}{2}}+e^{-\pi}-1+\frac{e^{-2 \pi}+1}{5}$

iii) To find other points of intersections of the normal and the curve, C.

Solve both eqautions simultaneously.

We know the equation of the normal is,

$y=-x+2e^{-\frac{\pi}{2}}$

and the curve C has equations defined parametrically by,

$x=e^{-t}, y=e^{-t}sin(t)$

Subsitute x and y into the normal and solve to obtain,

$e^{-t}sin(t)=-e^{-t}+2e^{-\frac{\pi}{2}}$

Solving for t gives,

t=-1.01, t=1.57

Need some with higher level mathematics integration question as attached in this post, specifically

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Ответ:

bri0521

29.02.2020

Mathematics

The 95% confidence interval for the difference would be given by (-1.776;0.376)

We are 95% confidence that the true mean difference is between $-1.776 \leq \mu_d \leq 0.376$ . Since the confidence interval contains the value 0, we don't have enough evidence to conclude that hypnotism appear to be effective in reducing pain.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let put some notation

x=test value before , y = test value after

x: 6.4 2.6 7.7 10.5 11.7 5.8 4.3 2.8

y: 6.7 2.4 7.4 8.1 8.6 6.4 3.9 2.7

The differences defined as d_i = y_i -x_i and we got:

d: 0.3, -0.2, -0.3, -2.4, -3.1, 0.6, -0.4, -0.1

We can calculate the mean and the deviation for the sample with the following formulas:

$\bar d=\frac{\sum_{i=1}^n d_i}{n}$

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1}$

$\bar d=-0.7$ represent the sample mean for the difference

$\mu_d$ population mean (variable of interest)

s_d =1.32 represent the sample standard deviation

n=8 represent the sample size

Confidence interval

The confidence interval for the mean is given by the following formula:

$\bar d \pm t_{\alpha/2} *\frac{s_d}{\sqrt{n}}$ (1)

In order to calculate the critical value t we need to find first the degrees of freedom, given by:

df=n-1=8-1=7

Since the Confidence is 0.95 or 95%, the value of alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,8)".And we see that $t_(\alpha/2)=2.306$

Now we have everything in order to replace into formula (1):

$-0.7-2.306\frac{1.32}{\sqrt{8}}=-1.776$

$-0.7+2.306\frac{1.32}{\sqrt{8}}=0.376$

So on this case the 95% confidence interval for the difference would be given by (-1.776;0.376)

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Need some with higher level mathematics integration question as attached in this post, specifically parts (iii) and (iv). you in advance to whoever who manages to ! : )

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