This question is in the Pre-Calculus sub-forum, are you allowed to use calculus?
i have completely forgot how to work this problem if anyone could give step by step way to solve. Thak you
What is the (closest) distance from the curve y=x^2 to point
x, y =(16,1 2) ?
Moderator edit: This thread was moved from the Pre-algebra and Algebra subforum to Pre-calculus subforum. In light of subsequent posts by the OP, the thread has now been moved to the Calculus subforum.
Yes.
And therefore the points on the parabola have the coordinates P(a, aČ) as Plato wrote.
The slope of the parabola at P has the value of 2a.
Since you need the perpendicular direction of the parabola's slope you have to use the property:
2 lines are perpendicular if their slopes satisfy the equation:
So if the first slope is 2a the 2nd slope is .
ok i got this far now
distance = sqrt[(16-x)^2 + (1/2 - y)^2]
= sqrt[(16-x)^2 + (1/2 -
x^2)^2]
= sqrt(256 - 32x + x^2 + 1/4 - x^2 + x^4)
=
(x^4 - 32x + 256.25)^(1/2)
now when i checked on another site the next step does not make any since i cannot figure out where they get the first part from.
d '(x) = (1/2)(4x^3 - 32)(x^4 - 32x + 256.25)^(-1/2)
set = 0
4x^3 - 32 = 0
where does the 4x^3 - 32 come from???
You were asked in post #2: Are you allowed to use calculus? Please answer this question.
And if you have not learned calculus then you should not be asking questions about a solution that uses calculus. (And more to the point, if you are asking the original question at another website and getting replies you don't understand, then that website is the place where you should be asking your question).
You have been given a non-calculus method in earlier posts (and people have spent their time giving it to you). If this is not the method you want to use, please say so. Otherwise, use the method given in those earlier posts and ask questions about it if you're still stuck.