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Mathematics: Post your doubts here!

D

DaniyalK

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farhan141 said:
Do u know how to convert a velocity time graph to distance time or vice versa? I'm rly facing issue in this
Click to expand...
It's not too bad as long as you understand that the v/t graph is the graph of the gradient of s/t and remember a few cases.
constant velocity is a straight diagonal line for displacement
straight diagonal line for velocity is increasing gradient graph for displacement
curved graph for velocity is increasing gradient at an increasing/decreasing rate for displacement, depending on the curve in the v/t graph

farhan141 said:
I have some doubts please clear.

- Why is 'c' ignored when integrating velocity to find disp?
Click to expand...
Because we are doing definite integration. The c cancels out.

Haya Ahmed said:
Won't they cut marks if I drew it in the wrong direction?
Click to expand...
There is no 'right' or 'wrong' direction. Positive can be any direction as long as you show motion in the opposite direction as negative.
 
H

hizirofaki

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c
DaniyalK said:
It's not too bad as long as you understand that the v/t graph is the graph of the gradient of s/t and remember a few cases.
constant velocity is a straight diagonal line for displacement
straight diagonal line for velocity is increasing gradient graph for displacement
curved graph for velocity is increasing gradient at an increasing/decreasing rate for displacement, depending on the curve in the v/t graph


Because we are doing definite integration. The c cancels out.


There is no 'right' or 'wrong' direction. Positive can be any direction as long as you show motion in the opposite direction as negative.
Click to expand...
an u plz sketch the raph for his 7 marks !!!! :confused:
 

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F

farhan141

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DaniyalK said:
It's not too bad as long as you understand that the v/t graph is the graph of the gradient of s/t and remember a few cases.
constant velocity is a straight diagonal line for displacement
straight diagonal line for velocity is increasing gradient graph for displacement
curved graph for velocity is increasing gradient at an increasing/decreasing rate for displacement, depending on the curve in the v/t graph


Because we are doing definite integration. The c cancels out.


There is no 'right' or 'wrong' direction. Positive can be any direction as long as you show motion in the opposite direction as negative.
Click to expand...


Can u give me an example to clear my doubt regarding definite integration and how C cancels?
 
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farhan141

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DaniyalK said:
Because we are doing definite integration. The c cancels out.


There is no 'right' or 'wrong' direction. Positive can be any direction as long as you show motion in the opposite direction as negative.
Click to expand...

Also solve my question I posted in previous page :)
 
D

DaniyalK

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farhan141 said:
Can u give me an example to clear my doubt regarding definite integration and how C cancels?
Click to expand...
suppose you integrate v and get this
S = 2t^2 - 3t + c
and you know s = 5 when t = 0
so c = 5
S = 2t^2 -3t + 5
now this is the term you will apply limits to so it becomes
Upper limit - lower limit
but whatever limits you use, the 5 will stay 5 because it has nothing to do with t
so whatever the limits are, you will always end up with 5 - 5 = 0 when you do upper limit - lower limit.
Therefore, we can just ignore it.
 
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DaniyalK

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farhan141 said:
Why isn't T2 20s? My values were 10 and 30 but when I inserted value of 21 in second derivative of velocity it was giving me positive so increasing? And when I put 19 it was giving me negative so decreasing?View attachment 53464
Click to expand...
Why are you putting them in the second derivative... Just take 1 derivative. That will give you acceleration. Velocity will be increasing when acceleration is positive and decreasing when acceleration is negative.
 
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farhan141

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DaniyalK said:
Why are you putting them in the second derivative... Just take 1 derivative. That will give you acceleration. Velocity will be increasing when acceleration is positive and decreasing when acceleration is negative.
Click to expand...
Isn't that u have to find d2y/dx2 to see if it is inc or dec. for that function.


Also, about the 'c'. I don't understand the limits thingy for displacement :3. Isnt it only for volume/area?
 
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DaniyalK

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farhan141 said:
Isn't that u have to find d2y/dx2 to see if it is inc or dec. for that function.


Also, about the 'c'. I don't understand the limits thingy for displacement :3. Isnt it only for volume/area?
Click to expand...
You may not know it but you ARE calculating the area. Area under v/t time graph is the displacement.
Or in that that example, think of it this way, suppose you want to know the displacement between t = 0 and t = 10.
So it will be displacement up till t = 10 - displacement at t = 0
so it will be S(10) - S(0) and the 5s will cancel.
At S(0), only the constant remains as anything containing t becomes 0. That's why you get the correct answer when you plug in just 1 value of t (10 in this case) in S, because the constants cancel. But it's important to understand that what you are actually doing is finding the area under the graph from t = 0 to t = X. But suppose if it asks you to find the displacement from t = 10 to t = 20. Then you subtract the displacement at 20 from the displacement at 10, in other words, you are finding the area under the v/t graph from 10 to 20.

d2y/dx2 tells you the nature of a stationary point. Increasing/decreasing is checked from the gradient, dy/dx.
 
F

farhan141

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DaniyalK said:
You may not know it but you ARE calculating the area. Area under v/t time graph is the displacement.
Or in that that example, think of it this way, suppose you want to know the displacement between t = 0 and t = 10.
So it will be displacement up till t = 10 - displacement at t = 0
so it will be S(10) - S(0) and the 5s will cancel.
At S(0), only the constant remains as anything containing t becomes 0. That's why you get the correct answer when you plug in just 1 value of t (10 in this case) in S, because the constants cancel. But it's important to understand that what you are actually doing is finding the area under the graph from t = 0 to t = X. But suppose if it asks you to find the displacement from t = 10 to t = 20. Then you subtract the displacement at 20 from the displacement at 10, in other words, you are finding the area under the v/t graph from 10 to 20.

d2y/dx2 tells you the nature of a stationary point. Increasing/decreasing is checked from the gradient, dy/dx.
Click to expand...

You are amazing thanks :)

Just one more thing. Does the same theory (ignoring c) apply for calculating distance?
 
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farhan141

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DaniyalK said:
The tensions AP and AQ are equal, so the angles on both sides of 3sqrt3 are the same. The angles are (180-90-30)/2 = 30
Now take the direction of the 3sqrt3 force as your plane and resolve forces along it. Look at the horizontal part.
Tcos30 + Tcos30 = 3sqrt3
T = 3
Click to expand...

Oh missed on the equal angles part, thanks :)
 
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