Solve Shortest Distance Point (0,c) from Parabola y=x^2

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The discussion focuses on finding the shortest distance from the point (0,c) to the parabola y=x^2 for values of c between 0 and 5. The initial solution provided is (c-0.25)^0.5, but there is confusion regarding its validity for c values less than 1/4. Participants highlight that if c is less than 1/4, the solution does not apply, and suggest checking the endpoints of the interval to determine the nearest point on the parabola. The importance of considering the entire range of c values is emphasized to ensure a complete understanding of the problem. Clarifying these conditions is crucial for accurately solving the distance problem.
gaganpreetsingh
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Hi I am stuck on this ques. although i have solved and got the ans but i am still facing a problem.
Find the shortest distance of the point (0,c) from the parabola y=x^2 where 0<= c <= 5
i got the ans right that is
(c-0.25)^0.5
but i have not been able to understand the use of the 0<= c <= 5
any help?
 
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any help out there?
 
Assuming that c> 1/4, I get that. However, notice that your solution makes no sense if c< 1/4! What is the nearest point on the parabola if c< 1/4? (Hint: check the endpoints of the interval of values for y.)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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