Hi! Can I get some help with this questions. I am very confused about how to approach it.
Find the point on the curve y=5x+4 closest to the point (0,6).
Thank you!
5 answers
y = 5x + 4 is a line, so I am not certain where the curve comes from... did you make a typo?
the closest point will be on a line passing through the given point, and perpendicular to the tangent of the curve
in this case, the curve is a straight line
the slope of the perpendicular is ... -1/5
find the equation of the line with a slope of -1/5, passing through (0,6)
the intersection with the original curve is the closest point
in this case, the curve is a straight line
the slope of the perpendicular is ... -1/5
find the equation of the line with a slope of -1/5, passing through (0,6)
the intersection with the original curve is the closest point
or, if you actually want to use some calculus, the distance d from any point (h,k) to a point on the line y=5x+4 is
d^2 = (h-x)^2 + (k-y)^2
= (h-x)^2 + (k-(5x+4))^2
= h^2-2hx+x^2 - (k^2 - 2(5x+4)k + (5x+4)^2)
= h^2-2hx+x^2 - k^2 - 10kx - 8k - 25x^2-40x-15
= (h^2-k^2) - (2h+10k+40)x - 24x^2
this distance is least when dd/dx=0, or
x = (h+5k+20)/24
or, when h = -5k-20
That is the line with slope -1/5 as shown above.
d^2 = (h-x)^2 + (k-y)^2
= (h-x)^2 + (k-(5x+4))^2
= h^2-2hx+x^2 - (k^2 - 2(5x+4)k + (5x+4)^2)
= h^2-2hx+x^2 - k^2 - 10kx - 8k - 25x^2-40x-15
= (h^2-k^2) - (2h+10k+40)x - 24x^2
this distance is least when dd/dx=0, or
x = (h+5k+20)/24
or, when h = -5k-20
That is the line with slope -1/5 as shown above.
MsPi: It does says it is a curve that is why I am confused on how to do it.
Scott: What do you mean by the intersection with the original curve is the closest point
Scott: What do you mean by the intersection with the original curve is the closest point
Steve: Thank you! And how would I calculate the point from there? I did that process before posting the question here and I couldn't figure out how to get the actual point
1 answer