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Thursday, March 19, 2009

Solved Problem on Partial Differentiation

Topic : Partial Differentiation

Problem : Solve the function by differentiation f(x,y) = xe^(x^2 y) at the intervals (1, ln 2)

Solution:

Following are the steps to differentiate a given function.

$f(x,y) = xe^{x^2y}\\\frac{\partial f}{\partial x}=? And \frac{\partial f}{\partial y}=?\\\frac{\partial f}{\partial x}_{(1,2)}=? And \frac{\partial f}{\partial y}_{(1,ln 2)}=?\\ f(x,y) = xe^{x^2y}<br />\\Consider \frac{\partial e^{x^2y}}{\partial x}\\\frac{\partial e^u}{\partial x} and u=x^2y\\\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\ as u=x^2y\\\frac{\partial e^u}{\partial u}=e^u=e^{x^2y}----------(a)\\\frac{\partial u}{\partial x}=2xy-------(b)\\ By Chain Rule\\\frac{\partial e^{x^2y}}{\partial x}=\frac{\partial e^u}{\partial u} . \frac{\partial u}{\partial x}\\=e^{x^2y} . 2xy where a = e^{x^2y} and b=2xy\\\frac{\partial f}{\partial x}=x \frac{\partial e^{x^2y}}{\partial x}+e^{x^2y}\frac{\partial x}{\partial x}\\=xe^{x^2y}(2xy)+e^{x^2y}(1)\\=e^{x^2y}[2x^2y+1]\\\frac{\partial f}{\partial x}=xe^{x^2y}(x^2)\\=x^3e^{x^2y}\\Therefore\frac{\partial f}{\partial x}_{(1,ln2)}=e^{1ln2}(i^2.ln2)=e^{ln2}=2ln2\\\frac{\partial f}{\partial y}_{(1,ln2)}=1^3 e^{1^2ln(2)}=e^{ln(2)}=2<br />$

Monday, March 16, 2009

Question on Probability of Students in a Class Room Under Different Contraints

Topic : Probability

Question : In a class of 72 students, 44 are girls and,of these, 12 are going to college. Of the 28 boys in the class, 9 are going to college. If a student is selected at random from the class, what is the possibility that the person chosen is going to college.

Solution :
12 girls college
9 girls college
No of students in a class = 72
Total number of students going to college = 12+9=21
probability that the student go to college = 21/72 = 7/24

Wednesday, March 11, 2009

Question on Simplifying an Exponent

Topic : Simplification

Question : Simplifying exponent (4xy)º

Answer :

(4xy)º = 1

According to law of Exponent (aº = 1), Any number or a variable to the indices zero will be equal to one.

Thursday, March 5, 2009

Question on Box Whisker Plot

Topic : Box Whisker Plot

Question : Explain Box Whisker Plot for the given sequence of numbers.

4,6,7,9,15,18,20,23,27,31,33,35

Solution :

4,6,7,9,15,18,20,23,27,31,33,35

Minimum value = 4
Maximum value = 35
The middle numbers are 18 and 20

Let Q2 is the middle term of whole sequence is 18 and 20
=(18+20)/2 = 38/2 = 19

Q2 = 19

And Q1 = (7+9)/2 = 16/2

Q1 = 8

Also Q3 = (27+31)/2 = 58/2

Q3 = 29

Friday, February 27, 2009

Problem on Calculating Percentage

Topic : Percentile Problem

Question : What is the % of 64 is 24?

Answer :

Let x%of 64 =24
So x/100*64=24
64x/100=24
0.64x=24 (divide both sides by 0.64)
x =24/0.64
x =37.5%

Hope the Answer is helpful.

Friday, February 6, 2009

simple Question on Combinations

Topic : Chess Board Problem

Question : Determine the number of rectangles that can be formed on a chess board.

Answer :

There are nine horizontal lines and there are nine vertical lines.To form a rectangle we need 2 vertical lines and 2 horizontal lines from each set.This can be done in 9C2 *9C2 ways= 36*36= 1296.

Hope the Answer is Helpful.

Word Problem about Distance travelled by a Camel with some Constraints

Topic : Word Problem on Distance

Question : A camel must travel 1000 miles across a desert to the nearest city. She has 3000 bananas but can only carry 1000 at a time. For every mile she walks, she needs to eat a banana. What is the maximum number of bananas she can transport to the city?

Answer :
Suppose the camel does this:
1. Pick up 1000 bananas.
2. Carry them one mile into the desert, eating 1 on the way.
3. coming back 1 mile and eating 1 banana;
4. pick up 1000 bananas and carry one mile into the desert , eating 1 banana;
5. coming back 1 mile and eating 1 banana;
6. pick up 1000 bananas and carry one mile into the desert , eating 1 banana;
At the end of this, there will be 2995 bananas, which are 999 miles from the city.
The camel will also be 999 miles from the city.Camel will need 5 bananas to transfer all the bananas 1 mile away…
5 as for 1st 1000 he'll go n come back then for another 1000 he'll go n come back but for last 1000 he'll only go as nothing will be left behind…
For second mile too it'll take 5 bananas…
Now it'll require 5 bananas per mile for total banana count > 2000and for banana count <= 2000 it'll eat 3 bananas…
and for <= 1000 it'll eat 1 banana….
so 1st 1000 bananas will be finished in 1000/5 = 200 Miles next 1000 in 1000/3 = 334 Miles…finally he'll have 1000 bananas and 466 miles to go…
At B camel will have 534 bananas

Hope the explanation helps you to solve the problem, if you have any queries leave your comments.