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:
\[x^3=\frac{500 }{ \pi}\]
Sunday, 27 January 2013
Tuesday, 15 January 2013
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.
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.