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Derivative - Max and Min application!
Alpha:
--- Quote from: sgvaibhav on January 24, 2010, 03:07:47 pm ---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 -.-
--- End quote ---
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~ :)
SGVaibhav:
--- Quote from: ~Alpha on January 24, 2010, 03:14:35 pm ---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~ :)
--- End quote ---
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...
astarmathsandphysics:
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
SGVaibhav:
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 ...
Alpha:
--- Quote from: sgvaibhav on January 24, 2010, 03:30:15 pm ---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 ...
--- End quote ---
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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