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Hey guys please I need help here in part (ii) what are we supposed to do ?! ...
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Mathematics: Post your doubts here!
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Could you tell me what is the Dy/dx expression?Hey guys please I need help here in part (ii) what are we supposed to do ?! ...
and I think they want you to prove perpendicular* of tangent and curve so gradient is Opposite ( M1 * M2 = -1 )
so check if both have the same gradient
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dy/dx is (4-x)/(y+2)
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You should substitute x=7 into the original equation; this should form a quadratic equation.... then after you found both values of y, you can substitute both values into dy/dx...Hey guys please I need help here in part (ii) what are we supposed to do ?! ...
Hope this helps!!
NO.
dy/dx = (4 - x)/(-x^2 + 8x + 2)
and you get the proof for part (ii) by this eq too.
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I got the same answer as Haya for dy/dx.... :|
dy/dx should be only in terms of x, you have y as well in that expression.
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Is it true? when I learned about the curves defined implicitly I think it's alright to include y in the dy/dx.... My math book also includes y in the dy/dx.... Also this website:
http://www.phengkimving.com/calc_of_one_real_var/04_more_on_the_der/04_02_impl_diffn.htm
http://www.solitaryroad.com/c359.html
Btw I used the A level mathematics Pure Mathematics 2 & 3, made by Hugh Neill and Douglas Quadling.
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implicit differenciation is abt includin y itself. thts y its called implicit
but at times in some word problems we've to eliminate the y so as to get dy/dx in terms of x or as per the question requirement
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TimBluesWin is right even the answer at the back of the book is like ours '' ... Thanks
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In this question guys I got the dy/dx which is "3" but now my doubt is how can we find the equation without having the y value ... !! ... the answer is
9y- 2x -e = 0 .. please answer ASAP
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Use the quotient rule.In this question guys I got the dy/dx which is "3" but now my doubt is how can we find the equation without having the y value ... !! ... the answer is9y- 2x -e = 0 .. please answer ASAP
Let u=x
u'=1
v=2+lnx
v'=1/x
dy/dx= (u'v-uv')/v^2
((2+lnx)-1)/((2+lnx)^2)
For the equation of the tangent, substitute the x=e to the equation to find the y, then substitute the e into dy/dx to find the gradient, the input your values to equation of line: y-y1=m(x-x1)
Hope this helps!
you can have the value of why by putting x = e in the eq of curve.In this question guys I got the dy/dx which is "3" but now my doubt is how can we find the equation without having the y value ... !! ... the answer is9y- 2x -e = 0 .. please answer ASAP
I dont know about that inplicit thing. I have just started my A levels, and have not studied differentiation in A levels yet.
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How to find K value ??? please with explaination ... the ans is -11
