230k views
1 vote
Find the point on the curve
y=√(5x+3) which is closest to the point (7,0)

User Domruf
by
8.1k points

1 Answer

5 votes

Answer: (4.5, 5.1)

Explanation:

The distance between two points (a, b) and (c, d) is:

D = √( (a - c)^2 + (b - d)^2)

Then if we want to find the point on the curve y = √(5*x + 3) that is closest to the point (7, 0) we need to minimize the distance:

D = √( (x - 7)^2 + (y - 0)^2)

D = √( (x - 7)^2 + (√(5*x + 3))^2)

Because we know that D is positive, minimizing D is the same than minimizing D^2

Then we can minimize:

D^2 = (x - 7)^2 + (5*x + 3)

D^2 = x^2 - 14*x + 21 + 5*x + 3

D^2 = x^2 - 9*x + 24

This is a quadratic equation with a positive leading coefficient, then the minimum of this function is at the vertex.

To find the vertex, we need to find the zero of the first derivative, this is:

(D^2)' = 2*x - 9

We need to solve:

0 = 2*x - 9

9 = 2*x

9/2 = x

4.5 = x

To find the correspondent y-value, we need to evaluate the curve in x = 4.5

y = √(5*4.5 + 3) = 5.1

Then the point is (4.5, 5.1)

This means that the point on the curve y = √(5*x + 3) which is closest to the point (7, 0) is the point (4.5, 5.1)

User Jimeux
by
8.3k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.