Question

The point (1,2) has the distance ________ from the line defined by y=−x3+1.
could somebody solve this problem

Answer 1

is the line \(\Large{y = -x^3+1}\)

Answer 2

yes

Answer 3

do you know the differential calculus?

Answer 4

no its my precalculus lecture's problem

Answer 5

ok!
Then I try to search for another method, please wait...

Answer 6

unless its a line, i don think there is a geometric way to approach it

Answer 7

d^2 = (x-k)^2 + (y-k)^2 is minimal

unless that setup simplifies to something simple ...

Answer 8

x-h not x-k

Answer 9

http://upload.wikimedia.org/math/c/a/e/cae82210ba48deb8ff6fb0134a90b36e.png
i think this is the formula to solve it but when i solve it was wrong

Answer 10

just wondering

d^2 = (x-1)^2 + (-x^3+1-2)^2

d^2 = x^2 -2x +1 + x^6+1+2x^3

d^2 = x^6 +2x^3 +x^2 -2x +2

unless the right side is a perfect square, i dont think this has a chance ...

Answer 11

where does your formula come from?

Answer 12

instructor

Answer 13

if all you are doing is trying to memorize formulas, then you are not being taught anything ... computer programs can determine the distances based on formulas

Answer 14

its not what i am doing i just suggest it

Answer 15

d^2 = (x-h)^2 + (y-k)^2

taking the derivative

2d d' = 2x' (x-h) + 2y' (y-k)

d d' = (x-h) + y' (y-k)

(x-h) + y' (y-k)
d' = ----------- = 0 when (x-h) + y'(y-k) = 0
d

Answer 16

i do not know calculus yet it an precalculus question

Answer 17

this is how i would approach it since its not a line

Answer 18

I found another method, nevertheless we have to know the meaning of the slope of a tangent line to a function as the first derivative of that function.

Answer 19

good luck

Answer 20

do you know the concept of first derivative of a function?

Answer 21

yes

Answer 22

ok! Now let's consider a generic point Q on your function y=-x^3+1, namely
\[Q = \left( {{x_0},{y_0}} \right)\]

Answer 23

or:
\[Q = \left( {{x_0},{y_0}} \right) = \left( {{x_0}, - x_0^3 + 1} \right)\]

Answer 24

is it ok?

Answer 25

Q belongs to our function f(x) = -x^3+1

Answer 26

yes i am ok with this solution

Answer 27

ok! Now I have to write the tangent line, to our function, at that point. The slope m, of that tangent line is:
\[m = f'\left( x \right) = - 3x_0^2\]

Answer 28

so the equation of our tangent line, is:
\[y - \left( { - x_0^3 + 1} \right) = - 3x_0^2\left( {x - {x_0}} \right)\]

Answer 29

please tell me, when I may continue