I have attached a picture of my problem regarding finding the point on a curve r(t) that is closest to another given point off of the curve. I have used the distance formula using the equation and given point in the i, j, and k directions. Why is it that when the derivative square of the distance is set to zero, the distance is minimized(second included screenshot)? Find thepoint P on thecurve r(t) that lies closest toPo and statethe distance between Po and P.r(t) = i+ 2tj + 2tk; Po(4,14,13) (? -4)2 + (2t – 14)2 + (2t – 13)2Distance =If the distance is minimized, then the square of the distance is also minimized. To find the minimum distance,differentiate the square of the distance with respect to t and find the value of t for which the result is zero.The square of the distance is (P - 4)2 + (2t – 14)2 + (2t – 13)2. Differentiate this expression with respect to t.+(2t - 14)2 + (2t – 13)2] = 413 – 108Now set the derivative equal to zero to find the value of t that minimizes the distance.4t3 - 108413= 108t = 3

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Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN:9780547587776

Author:HOLT MCDOUGAL

Publisher:HOLT MCDOUGAL

Chapter9: Real Numbers And Right Triangles

Section: Chapter Questions

Problem 28CR

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I have attached a picture of my problem regarding finding the point on a curve r(t) that is closest to another given point off of the curve. I have used the distance formula using the equation and given point in the i, j, and k directions. Why is it that when the derivative square of the distance is set to zero, the distance is minimized(second included screenshot)?

Find the
point P on the
curve r(t) that lies closest to
Po and state
the distance between Po and P.
r(t) = i+ 2tj + 2tk; Po(4,14,13)

(? -4)2 + (2t – 14)2 + (2t – 13)2
Distance =
If the distance is minimized, then the square of the distance is also minimized. To find the minimum distance,
differentiate the square of the distance with respect to t and find the value of t for which the result is zero.
The square of the distance is (P - 4)2 + (2t – 14)2 + (2t – 13)2. Differentiate this expression with respect to t.
+(2t - 14)2 + (2t – 13)2] = 413 – 108
Now set the derivative equal to zero to find the value of t that minimizes the distance.
4t3 - 108
413
= 108
t = 3

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