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Mathematics: Post your doubts here!

whitecorp

whitecorp

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Soldier313 said:
whitecorp and others
need some help here integrating:
find the volume of the solid formed when the graph of y = (1/x*2) between x=1 and x=2 is rotated about the x - axis.
answer is: 2 Π ln2
Click to expand...

Is that y=1/(x^2) ? Because if it is then I am positive the answer you provided is incorrect.
 
Soldier313

Soldier313

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whitecorp said:
Is that y=1/(x^2) ? Because if it is then I am positive the answer you provided is incorrect.
Click to expand...
yes it is.....the textbook says the answer is 2.pi.ln2....what answer did you get...and how did you arrive at it? the textbook might be wrong
 
whitecorp

whitecorp

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Taiyaba said:
whitecorp plz help me with second one :(
Click to expand...

The series 1+3+5+...........+(2n-1) constitutes an arithmetic progression (AP) that has n terms in all, first term of value 1 and common difference =2.
Then we can simply use formula for sum to n terms, which is given by (number of terms divided by 2)*(sum of first term and last term)
=n/2 * (1+2n-1) = n/2 * (2n) =n^2 (shown)

Hope this helps. Peace.
 
Taiyaba

Taiyaba

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whitecorp said:
The series 1+3+5+...........+(2n-1) constitutes an arithmetic progression (AP) that has n terms in all, first term of value 1 and common difference =2.
Then we can simply use formula for sum to n terms, which is given by (number of terms divided by 2)*(sum of first term and last term)
=n/2 * (1+2n-1) = n/2 * (2n) =n^2 (shown)

Hope this helps. Peace.
Click to expand...
Thnx :) bt hw do v show that it is always a perfect square?
 
VelaneDeBeaute

VelaneDeBeaute

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bamteck
For Question#3, we apply the parallelogram rule and hence construct a disgram like this is mind. Then take one of the two triangles and apply sine rule to them. If you consider the first triangle in the diagram below, then Q/sin40 = 12/sin60. Solve this to get the value of Q directly as 8.91
Q_3.JPG
 
VelaneDeBeaute

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@bamtech For Q#4, i've resolved the 5N force in the direction of the 40cm string. The force will hence be 5/cos A if A is the angle formed b/w the 5N force and its component. Applying sine rule to the triangle and using the given lengths, we first find the angle formed b/w PS and AS which is 36.9. Then considering the right-angled triangle formed, you just subtract 90+36.9 out of 180 to find the angle between PS and its vertical. The angle is equal to A(as its vertically opp.) and when put into 5/cos A gives you the tension in the string PS.
Q4.JPG
 
bamteck

bamteck

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VelaneDeBeaute said:
@bamtech For Q#4, i've resolved the 5N force in the direction of the 40cm string. The force will hence be 5/cos A if A is the angle formed b/w the 5N force and its component. Applying sine rule to the triangle and using the given lengths, we first find the angle formed b/w PS and AS which is 36.9. Then considering the right-angled triangle formed, you just subtract 90+36.9 out of 180 to find the angle between PS and its vertical. The angle is equal to A(as its vertically opp.) and when put into 5/cos A gives you the tension in the string PS.
View attachment 16174
Click to expand...

thanks a lot :)
much appreciated ! :)
 
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