# Thread: distance for curve to point help

1. ## distance for curve to point help

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.

2. This question is in the Pre-Calculus sub-forum, are you allowed to use calculus?

3. Originally Posted by jessmari86
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) ?
I agree that you really need calculus for this, (see reply #2)
But the shortest distance is a perpendicular distance.
Find a point on $y=x^2$, $(a,a^2)$ so that the slope $\dfrac{12-a^2}{16-a}=\dfrac{-1}{2a}$.

Now you need calculus to know why that works.
BTW: the point is in quadrant I.

4. ok so where does the a come from? is that supposed to represent any point on that is the curve?

5. Originally Posted by jessmari86
ok so where does the a come from? is that supposed to represent any point on that is the curve?
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 $l_1, l_2$ are perpendicular if their slopes $m_1,m_2$ satisfy the equation: $m_1 \cdot m_2 = -1$

So if the first slope is 2a the 2nd slope is $-\frac1{2a}$.

6. Originally Posted by earboth
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 $l_1, l_2$ are perpendicular if their slopes $m_1,m_2$ satisfy the equation: $m_1 \cdot m_2 = -1$

So if the first slope is 2a the 2nd slope is $-\frac1{2a}$.

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???

7. Originally Posted by jessmari86
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???

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.

8. Originally Posted by mr fantastic

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.
yes calculus can be used.

9. Originally Posted by jessmari86
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) Mr F says: This is the derivative. You get it using the chain rule.

set = 0

4x^3 - 32 = 0 Mr F says: The derivative is equated to zero to find the x-coordinates of the stationary points of the distance function. Note: If a/b = 0 then a = 0. That is, since the derivative is equal to zero, the numerator of the derivative must be equal to zero.

where does the 4x^3 - 32 come from???
There is an extensive amount of material you need to review if you are to understand questions like this one. I suggest you take the time (perhaps a week or 2 ...) to so so. Then show what progress you have made with the question and, if necessary, ask for more help.