Find Min Distance from x=20 to y=3√(x-15)

[buffgilville]

Find the minimum distance from the point where x = 20 along the x-axis to the curve given by:
y= 3 * square root of (x-15)

My prof. has not gone over this yet, so can someone help me?

 

[Spectre5]

create a vector, measured from the origin that locates any point from the x = 2 mark to the line. Then find the magnitude of that vector (in terms of x). Differentiate that answer and set it equal to zero. Find your solution. Check to make sure it is a mininum and not a maximum.

Then you have your x value. Plug that value into the equation for the maginute of the vector, that is the distance

 

[wais]

To find the minimum distance from the point where x = 20 to the curve y = 3√(x-15), we can use the concept of derivatives. The minimum distance occurs when the tangent line to the curve at that point is perpendicular to the x-axis. This means that the derivative of the curve at that point must be equal to 0.

First, let's find the derivative of the curve y = 3√(x-15):

y' = (3/2)*(x-15)^(-1/2)

Now, we can set this equal to 0 and solve for x:

(3/2)*(x-15)^(-1/2) = 0

(x-15)^(-1/2) = 0

x-15 = 0

x = 15

This means that the minimum distance occurs at x = 15. To find the minimum distance, we can plug this value into the equation for y:

y = 3√(15-15)

y = 0

Therefore, the minimum distance from the point where x = 20 to the curve y = 3√(x-15) is 0. This makes sense because at x = 15, the point on the curve and the point on the x-axis are the same, so the distance between them is 0.