Pure Maths 31

TaffsAsLevel

TaffsAsLevel

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I think it is mixed of vdu/dx-udv/dx / v^2 and implicit after that by substituting y with the given equation but then I also got stuck the mark scheme not giving the final form of the expression
 
TaffsAsLevel

TaffsAsLevel

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Oh wait the final expression is given itself in the question, the gradient. I'll check my work
 
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Oh yes, I got the answer, first do what I said and then get dy/dx as subject after implicit differentiation, then simplify it, you get to the point where you have to multiply by (1+x)^(1/2) / (1+x)^(1/2) then you will see difference of two squares, there bingo good luck
 
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surinr

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is there anyway you can give me a detail explanation. where did the implicit differentiation come from?
 
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differentiate both sides, one by quotient and the other half by implicit because it contains y^2
 
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Arjun Dhanak

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Hi. I need to know when we use vector cross product and what the reason for using it is. What is the outcome of doing vector product?
Thanks in advance.
-Arjun
 
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surinr

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Arjun Dhanak said:
Hi. I need to know when we use vector cross product and what the reason for using it is. What is the outcome of doing vector product?
Thanks in advance.
-Arjun
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im not too strong with the vectors but the cross product gives us a vector which is perpendicular to the plane or several vectors on the plane
 
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