Tangent to a parametrized curve

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The discussion focuses on finding the equation of the tangent line to a parametrized curve defined by x=t^2-6 and y=t^3+3t at t=3. The coordinates at t=3 are calculated as x(3)=3 and y(3)=36. The slope of the tangent line is determined using the derivative ratio, yielding m=5. The equation of the tangent line is derived as y=5(x-3)+36, which simplifies to y=5x+21. There is a note regarding the format of the final equation, suggesting it does not meet the specified requirements.
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Homework Statement
Given ##x=t^2-6## and ##y=t^3+3t##, what is the equation of the tangent to the curve when ##t=3##.
Relevant Equations
##m=y'(t)/x'(t)##
##x(3)=9-6=3##, ##y(3)=27+9=36##.
##\frac{y'(3)}{x'(3)}=\frac{3\times9+3}{2\times3}=\frac{30}{6}=5##.
##y=5(x-3)+36=5x+21##.
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archaic said:
Homework Statement:: Given ##x=t^2-6## and ##y=t^3+3t##, what is the equation of the tangent to the curve when ##t=3##.
Relevant Equations:: ##m=y'(t)/x'(t)##

##x(3)=9-6=3##, ##y(3)=27+9=36##.
##\frac{y'(3)}{x'(3)}=\frac{3\times9+3}{2\times3}=\frac{30}{6}=5##.
##y=5(x-3)+36=5x+21##.
View attachment 260822
I can only suggest the format is not as required. It says to type an equation, but 5x+21 is not such.
 
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haruspex said:
I can only suggest the format is not as required. It says to type an equation, but 5x+21 is not such.
o:)
 
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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