AP Calculus:
Find the point on the graph of the function that is closest to the given point.
f(x)=(x+1)^2 Point (5,3)

Question

AP Calculus:
Find the point on the graph of the function that is closest to the given point.
f(x)=(x+1)^2 Point (5,3)

anonymous · Answer

i dont know what to do

jennychan12 · Answer

take the derivative of the function. then plug in 5

anonymous · Answer

find the normal to this curve and constrain it to pass the following point and point on the curve and this point dist is the answer

anonymous · Answer

ok the derivative of the function is: $$2(x+1) \times1$$

anonymous · Answer

@RajshikharGupta huh?

jennychan12 · Answer

ok, then plug in 5

anonymous · Answer

$$2(5+1) \times 1$$

jennychan12 · Answer

yep :)

anonymous · Answer

then what

eyust707 · Answer

The function does not pass through that point. We want the line connecting our point points to be perpendicular to the slope f'(x).

jennychan12 · Answer

you have a point and a slope.

anonymous · Answer

okay..

anonymous · Answer

how did u get m=10?

jennychan12 · Answer

my bad sorry.
point slope form (5,3) m =12

anonymous · Answer

how did u get m=12?

anonymous · Answer

i didnt knowwere supposed to use point slope form on these types of problems.

jennychan12 · Answer

the derivative is the slope. your calc teacher should've told you that...
u dont have to; it's just easier

eyust707 · Answer

@jennychan12 that is not correct

We have a function and some point which does not lie on the function |dw:1354854114157:dw|
Which point on the function is closest to the other point in space?

anonymous · Answer

NO!NO! wait MAN what r u doing it is tangent not normal 
shortest distances r along normal

eyust707 · Answer

I've never actually done this but I can' see it being that hard... I'll derive it if no one else does

anonymous · Answer

jennychan12 is wrong

anonymous · Answer

$$d=\sqrt{(x-5)^2 + ((x+1)^2-3)^2}$$
$$d'=\frac{ 1 }{ 2 } ((x+5)^2 + (x^2 + 2x -2)^2)^\frac{ -1 }{ 2 } \times \left[ 2(x+5) \times 1 + 2(x^2 + 2x  -2) \right]$$

agent0smith · Answer

One way to do this is using the distance formula. Then you find the minimum of that function, by differentiating it and finding where the derivative is equal to zero.

anonymous · Answer

thats what my teacher showed me ^

anonymous · Answer

just in place of 12
do-1/12 then evry thing after that is correct

anonymous · Answer

@agent0smith can u explain how he got that by using Distance formula? ^^"

jennychan12 · Answer

sorry. i thought it was a tangent line problem.

anonymous · Answer

@agent0smith can u explain to me how he got that?

agent0smith · Answer

He's using the distance formula, and entering the points x2 - x1 and y2 - y1 into the formula. x1 = 5 and y1 = 3, and y2 you find in terms of x, by re-arranging your original f(x) equation. I'll write it out a little better, using the equation writer, so you can see.

jennychan12 · Answer

just saying, no need to take derivative of the whole square root problem.
just take the derivative of the inside. it'll give u the same answer. don't believe me? try it yourself.

anonymous · Answer

@jennychan12 yeah but I wanna try the method that my teacher tried to show me

agent0smith · Answer

Jenny is right, just minimize the d^2, no need for the square root. Since d^2 is a minimum when d is a minimum. distance formula is here: http://math.about.com/library/bldistance.htm $$x _{1}$$ is just x, that's the point we want to find.$$x _{2}$$ is 5. $$y _{2}$$ is (x+1)^2 and $$y _{1}$$ is 3.