Tangent to a parametrized curve

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SUMMARY

The discussion focuses on finding the equation of the tangent to the parametrized curve defined by the equations x(t) = t² - 6 and y(t) = t³ + 3t at the point where t = 3. The calculations yield the point (3, 36) and a slope of 5, resulting in the tangent line equation y = 5x + 21. The relevant derivative formula used is m = y'(t)/x'(t), which is applied to determine the slope at the specified parameter value.

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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:)
 

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