James post on Y!: A2 Maths M1 Question Help Needed!?

Question:

So i got this question and i managed to do part i and ii however i'm struggling to do part iii, can someone show me how to do this.
A particle P is projected vertically upwards, from horizontal ground, with speed 8.4 m s−1

(i) Show that the greatest height above the ground reached by P is 3.6 m. [3]A particle Q is projected vertically upwards, from a point 2 m above the ground, with speed um s−1

The greatest height above the ground reached by Q is also 3.6 m.(ii) Find the value of u. [2]It is given that P and Q are projected simultaneously.(iii) Show that, at the instant when P and Q are at the same height, the particles have the same speedand are moving in opposite directions. [6]

Solution:
 Take the ground level as being the zero for displacement and measure displacements as positive upwards.

(i) The equation of motion for P is: \( a = s'' = -g \) m/s^2.

Integrating once we get: \( v = s' = -g t + 8.4 \) m/s

and again:\( s = - (g t^2)/2 + 8.4 t \)m

The maximum height is achieved when \(v=0\), so occurs at \(t_{mx}= 8.4/g \) seconds, substituting this into the displacement equation we get the maximum displacement (maximum height) is:

\( \phantom{11} s_{mx} = - (g (8.4)^2)/2 + 8.4 (8.4/g) = (8.4)^2 / (2g) \approx 3.59 \approx 3.6 \) m,

(ii) For Q we have:\( a=-g,\  v = -gt +u\) and \(s= -(g t^2)/2 + ut + 2 \).

Q reaches its maximum height when \(v = 0\), so at \(t_{mx} = u/g\) s. At this time the displacement is:

\( \phantom{11} s_{mx} = - g (u/g)^2 /2 +u^2/g +2 = u^2/(2g) +2 = 3.6\) m

So now we can solve for \(u\).

(iii) You now have the displacement equations for both P and Q and the two particles are at the same height at the time when these two displacements are equal. So equate the displacements which will give you a quadratic in t. You want the positive root of this which gives the time when they are at the same height. Now plug this time into the two velocity equations ...

Lagrangian Interpolation

Given a set of points \( (x_1,y_1), (x_2,y_2) \dots (x_n,y_n)\) we want to find a polynomial of degree \(n-1\) which passes through all of the given points.

We start by observing that:

\[P_j(x) = y_j\frac{ \prod_{i=1, i\ne j}^n (x-x_i)}{\prod_{i=1, i\ne j}^n (x_j-x_i)}\]
is a polynomial of degree \(n-1\) which is zero for all of the \(x_i\)'s except for \(i=j\) where it takes the value \(y_j\). This is by construction since the polynomial is set up to have roots \(x_i,\ i=1..j-1,j+1, ...n\), and to take the value \(y_j\) at \(x_j\)

Now consider the polynomial:
\[P(x)=\sum_{j=1}^n P_j(x)\]
This is of degree \(n-1\) and only the \(k\)-th term is non-zero at \(x_k\) where it takes the value \(y_k\) and so \(P(x_k)=y_k\). Hence \(P(x)\) is a polynomial of degree \(n-1\) which passes through all the points in our given set.

Example:
Take the points: \( (1,2), (2,1), (4,3), (7,5) \), then our polynomials are:
\[ \begin{aligned} P_1(x)&=-\frac{(x-2)(x-4)(x-7)}{9}\\ P_2(x)&=\phantom{-}\frac{(x-1)(x-4)(x-7)}{10}\\P_3(x)&=-\frac{(x-1)(x-2)(x-7)}{6}\\P_4(x)&=\phantom{-}\frac{(x-1)(x-2)(x-4)}{18}\end{aligned}\]
Then combining these and simplifying:
\[P(x)=P_1(x)+P_2(x)+P_3(x)+P_4(x) =-\frac{11x^3-137x^2+424x-478}{90}\]

Samantha's EDEXCEL C1 AS question from Y!A

I need help with this maths questions from edexcel as level maths, core maths 1, Ex 7I Q13?


If you don't have the book then the question is:

The total surface area of a cylinder \(A\)cm2 with a fixed volume of \(1000\) cubic cm is given by the formula \(A = 2\pi x^2 + 2000/x\), where \(x\) cm is the radius. Show that when the rate of change of the area with respect to the radius is zero, \(x^3 = 500/ \pi\)

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Solution:


The expression "when the rate of change of the area with respect to the radius is zero" means when \( \frac{dA}{dx}=0\).

So differentiate \(A\) with respect to \(x\):

\[\frac{dA}{dx} = 4 \pi x + \frac{2000 (-1) }{ x^2}\]
Now we want to find \(x\) so that \(\frac{dA}{dx} = 0\) so you need to solve

\[4 \pi x + \frac{2000 (-1) }{ x^2} = 0\]
for \(x\). This rearranges to:

\[4 \pi x = . \frac{2000 }{ x^2}\]
or:

\[x^3=\frac{500 }{ \pi}\]

Mean and Standard Deviation of two uniform distribution functions?

T1's Question from Y!Answers:

I have two uniform distribution functions, one between X=2 and X=8, and another between X=9 and X=10.

The probability of any number is 1/7. Before X=2, between X=8 and X=9 and after X=10, the probability is 0.

I need to find the mean and standard deviation of X.

Answer:

The pdf of your distribution is:

\[f(x) = \left\{ \begin{array}{ll} 0,& x\lt 2\\
1/7,& 2 \le x \le 8 \\
0,& 8 \lt x \lt 9\\
1/7,& 9 \le x \le 10\\
0,& 10 \lt x \\ \end{array}\right. \]

The mean is:
\[ \mu = E(x) = \int_{-infty}^{infty} x f(x) dx\]
which we can decompose into a sum of integrals over intervals where \(f(x)\) is non-zero:
\[E(x) = \int_2^8 \frac{x}{7} dx  + \int_9^{10} \frac{x}{7} dx \]
so:
\[ \mu = E(x) = (1/7) (8^2/2 - 2^2/2) + (1/7) (10^2/2 - 9^2/2) = 5.5\]

The variance is \(E( (x-\mu)^2 ) = E(x^2) - \mu^2\).

Now use the same method as before to evaluate \(E(x^2)\) by decomposing the integral into the sum of integrals over the intervals where \(f(x)\) is non-zero.

The standard deviation is then the square root of the variance.