Find the point on the line y = 2x + 3 that is closest to the origin
how to solve this using calculus?
You need to come up with a function to minimize. any ideas?
optimization? all i know about it is that i have to take the first derivative then equate to 0
If you took any point on the plane (x,y), what is the distance from that point to the origin?
hmm?
Taking the derivative and setting it equal to 0 is the easy part of this problem. Figuring out what you will take the derivative of is the hard part. Which is what im trying to help you figure out.
so you're telling me i won't take the derivative of 2x + 3?
nope, thats not the right function.
interesting
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What is the distance from a point (x,y) to the origin?
my brain hurts here....
Right triangles . . .
i have no clue what either of you are implying
i can see from @joemath314159 's drawing that it will be the point where the slope is undefined i suppose...
the question is "what point is closest to the origin", so you have to ask yourself, what is the distance to the origin?
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since the point perpendicular to the origin is the one closest to the origin
do then it would be the y-intercept of 2x + 3....which would be (0,3)
correct me if i'm wrong...
That would be my intuitive guess, but you asked for the calculus solution. It's not necessarily that it is the point perpendicular, but is the straight line distance from the point to the line.
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oh yeah...what i did wasn't perpendicular
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