IGCSE/GCSE/O & A Level/IB/University Student Forum
Teachers and Students => Universities => Topic started by: SGVaibhav on January 08, 2010, 09:53:06 am
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Question asked in the picture below!
THanks a lot for help!!
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I dont know how to find the closest point.
However, about the formula, Both ur suggested formulas are correct.
This is because its *squared*
lets say if x =10
first formula : (18-10)^2 = 64
second formula : (10-18)^2 = 64
hope that helped
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I can't view the pic. ???
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alpha, its weird that u cannot see the pic ???
astar, i am getting different answers everytime i solve this, :( :( :(
can u please solve it with some working.
i dont understand what to find out after getting the derivative :(
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I really can't see... I've tried all that I could. ??? :-[
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try now
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where ru astar im dying here !!
lol im so dependent on him..
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Closest to the point >>> Line joining (18, 0) to the required point must be perpendicular to the point on the curve.
>>> Grad. at point on curve = Inverse sign, take reciprocal of grad. at point on curve
Let point be (x, y).
dy/dx = 2x
(x, y) (18, 0)
(y - 0)/ ( x - 18) = - 1/ 2x
y = - 1/ 2x (x- 18)
On line which passes th. (x, y), y = x^2 (eq. of curve).
x^ 2 = -1/ 2x (x- 18)
2x^3 = -x + 18
Can you complete the rest...?
Am not sure of the answer, it's a trial only. :-\
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i did not understand the method here...
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(y-0)^2=(0-y)^2 so both are the same.
d^2=(x-18)^2+(y-0)^2 and y=x^2
d^2=x^2-36x+324+x^4
differentiate to get 2x-36+4x^3=0
2x^3+x-18=0
(x-2)(2x^2+4x+9)=0
x=2 cos the quadratic has no solutions.
and y=2^2=4
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i did not understand the method here...
Use of gradients of tangents and perpendicular lines... :-\ Nope?
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i finally decided to ask my sir for help, but he is on vacation to malaysia.
i will ask my friends, because they use some different method.
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Sorry, was in a hurry...
Okay, lemme try to explain what I used.
Point on curve closest to (18, 0).
In general, when you're required to find the closest distance, you look for a line perpendicular to your point.
If I give you a set of parallel lines, the closest distance between them is the perpendicular height.
That's what I used. The tangent to the point (you are required to find) on the curve must be perpendicular to the line joining the point required (on the curve) and (18, 0).
I tried to draw it... Don't laugh, did that too in a hurry. :P ;D
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lol thanks for the effort , the drawing is good :D :P
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lol thanks for the effort , the drawing is good :D :P
Come-well! :D
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Hey you got the answer??
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hmm @alpha..
i plan to ask my friend today :P, not yet done ahhaha
astar, can u answer this question for me
1) we have a equation of a curve , and equation of the tangent line.
so is the derivative of the equation of the curve = derivative of the equation of the tangent line at intersection ?? (if this question sounds confusing, then i will paste the main question) ....
2) MAIN QUESTION - Find the value of k, if the line y = 2x is tangent to the curve y = x^2 + k...
this question annoys me =.= -.- :-\
thanks a lot :D
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Hope this time, it's clear... ???
Attachment, download.
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Hope this time, it's clear... ???
Attachment, download.
hmm
going in my head
not read completely, yet
because im watching TV + reading ur solution + eating + finding something :D
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hmm
going in my head
not read completely, yet
because im watching TV + reading ur solution + eating + finding something :D
LOL :P
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Hope this time, it's clear... ???
Attachment, download.
understood :D
thanks for ur efforts and lots of help
ur way is more logical actually and its better
we were taught in some weird way -.-
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understood :D
thanks for ur efforts and lots of help
ur way is more logical actually and its better
we were taught in some weird way -.-
Welcome. :) Nice that I could help. Till am alive. ;)
You know, teachers tend to hurry with the syllabus, that's why they don't get time to explain everything in detail. Try to read as much as you can.
Good Luck!
And yes, use discriminant for Question 2 above. Discriminant = 0.
I gotta go now, see ya~ :)
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Welcome. :) Nice that I could help. Till am alive. ;)
You know, teachers tend to hurry with the syllabus, that's why they don't get time to explain everything in detail. Try to read as much as you can.
Good Luck!
And yes, use discriminant for Question 2 above. Discriminant = 0.
I gotta go now, see ya~ :)
what is discriminant??
leave the first question
my main is this one only, confusing me -.-
MAIN QUESTION - Find the value of k, if the line y = 2x is tangent to the curve y = x^2 + k...
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ax^2+bx+c=0
discriminant is b^2-4ac
y=2x os tangent to y=x^2+k so the only meet at one point
2x=x^2+k has only one root
so x^2-2x+k=0 discriminant =0 for only one root
b^2-4ac=(-2)^2-4*1*k==0 so 4-4k=0 so k=1
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i was told in this way
derivative of curve = derivate of tangent line at intersection
y' = 2x (derivative of curve)
y = 2 (derivative of tangent line)
2x = 2
x = 1
when x = 1 ,, y = 2(x) --> y = 2(1) --> y = 2
tangent point (1,2)
y = x^2 + k
2 = 1^2 + k
2 = 1 + k
k = 1
that was the way i was told ???
lol i wanted to know what my sir did actually ...
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i was told in this way
derivative of curve = derivate of tangent line at intersection
y' = 2x (derivative of curve)
y = 2 (derivative of tangent line)
2x = 2
x = 1
when x = 1 ,, y = 2(x) --> y = 2(1) --> y = 2
tangent point (1,2)
y = x^2 + k
2 = 1^2 + k
2 = 1 + k
k = 1
that was the way i was told ???
lol i wanted to know what my sir did actually ...
Your teacher used the properties of gradients. You are told in the question that the line is a tangent to the curve. At one point on a curve, there can only be ONE tangent. By logic, the tangent on the curve at the point of contact with the line y = 2x, imperatively has to be the line y = 2x itself.
Your teacher found the gradient of the line y = 2x, and the gradient of the curve (that is, the tangent) at the point he assumes to be (x, y), where the line and the curve touch. Since the line y = 2x is itself the tangent, both gradients must be the same. That's what your teacher compared to get the value of x, then y. And finally, he replaces to get k.
That's a simpler method. ;)
Discriminant, as Astar said, is b^2 - 4ac. Refer to your Quadratic Formula for finding roots, the value of b^2 - 4ac, the discriminant, determines whether you will have real and distinct roots (b^2 - 4ac > 0), equal ones (b^2 - 4ac = 0), or imaginary ones (b^2 - 4ac < 0).
A tangent touches a curve at ONLY ONE point. Naturally, you should get only one value of x on solving both equations. Again referring to the Quadratic Formula, x will have only one value if b^2 - 4ac = 0, i.e. x= -b/2a.
That's the method Mr Paul used.
:)
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yeah, for both
exam got over yesterday xD, and i want to get good marks this time ...
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yeah, for both
exam got over yesterday xD, and i want to get good marks this time ...
Oh sorry. :( I'm one day late. I didn't know. Was busy too with everything, in general. -_-
Hope you get good results this time. :)
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Oh sorry. :( I'm one day late. I didn't know. Was busy too with everything, in general. -_-
Hope you get good results this time. :)
nah no sorry
i got the answer also
and if u could not tell, np, no need to say sorry >:(
there r other people to annoy
i remeber one weird thing
there was one question in midterm, for derivatives related rates..
i did not know how to solve it ...
i found the paper hours before the exams, and i asked a person ( a topper) to solve it for me and to explain it. that person taught the complete concept to me , and i understood
and in the exam , almost the same question with the same concept came.
guess what, i got the correct and and the other person got it wrong ... :o
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:)
Such "weird things" often happen. :-[
With everyone. When people ask you to explain just before exams, you yourself get tensed. I can understand perfectly why. -_-