anonymous
  • anonymous
order: Find the gradient of the normal to the curve, at the specified value of t. \[x=\cos^3t~~~~y=\sin^3t~~~~~~when ~~~t={1\over6}\pi\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@dumbcow
anonymous
  • anonymous
x'= -3cos^2tsint??? y'= 3sin^2tcost???
dumbcow
  • dumbcow
yes :)

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anonymous
  • anonymous
\[3\sin^2tcost \over -3\cos^2tsint\]
anonymous
  • anonymous
\[-sint \over cost\]?
dumbcow
  • dumbcow
good
anonymous
  • anonymous
-tant?
dumbcow
  • dumbcow
yes either way would work since you still have to evaluate at t=pi/6
anonymous
  • anonymous
Ok. let me evaluate. How can you do this using pi ? I always convert to decimals...?
dumbcow
  • dumbcow
pi/6 converted to degrees is 180/6 or 30
anonymous
  • anonymous
OK -tan30= -0.57735 m(-0.57735)=-1 m=1.73205 How do I convert this into sqrt?
dumbcow
  • dumbcow
looks good to leave it in radicals -tan(30) = -1/sqrt3 m = sqrt3
anonymous
  • anonymous
Is there some sort of formula for radicals?
dumbcow
  • dumbcow
no not really, well the unit circle since 30 degrees is on there sin(30) = 1/2 cos(30) = sqrt3/2 -tan(30) = -1/2 * 2/sqrt3 = -1/sqrt3
anonymous
  • anonymous
Hm Ok. Thank you! In exams, do they regard radicals over decimals? Or do you not know?
dumbcow
  • dumbcow
depends on the teacher or policies of school radical is considered more exact while decimal are approximate answers
anonymous
  • anonymous
Hm Ok. It's external examinations.
anonymous
  • anonymous
is this trigonometry? @dum
anonymous
  • anonymous
@dumbcow
anonymous
  • anonymous
Trig, yes and parametric equations
dumbcow
  • dumbcow
yeah ^^
anonymous
  • anonymous
ooh, more advance,.
anonymous
  • anonymous
Thanks for helping!

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