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        What is the least distance between (0,12) and (a(t^2),2at) where a is constant and t is variable.
one month ago

Sukant Kumar
27 Points
							Let the distance between the given points be $D$, then we know that $D=\sqrt{(at^2)^2 + (2at-12)^2}$ $D=\sqrt{(a^2t^4) + (4a^2t^2+144-48at)}$ $D=\sqrt{a^2t^4+ 4a^2t^2-48at+144}$ Let, as when f(t) will be minimum and then D will also be minimum, $f(t)=a^2t^4+4a^2t^2-48at$ $f'(t)=4a^2t^3+8a^2t-48a=a(4at^3+8at-48)$ $f''(t)=a(12at^2+8a) > 0, \forall, t \in R$, hence f’(t) is monotonically increasing. Minimum of f(t) is when f(t) reaches zero. The answer thus depends on the value of $a$ and cannot be resolved easily in terms of a. So let us say that at t1 the value of f(t) is zero and f(t) is minimum, right? Then, $f'(t_1)=4at^3+8at-48=0 \rightarrow f(t) = a^2t_1^4+4a^2t_1^2-48at_1=at_1(at_1^3)+4a^2t_1^2-48at_1=at_1(\frac{48-8at_1}{4})+4a^2t_1^2-48at_1=2a^2t_1^2-36at_1=2at_1(at_1-18))$Hence, the expression doesn’t resolve to a constant value and the minimum distance will not have a uniform expression for different values of a.

27 days ago
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### Course Features

• 53 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions