order: Find the gradient of the normal to the curve, at the specified value of t.
$$x=\cos^3t~~~~y=\sin^3t~~~~~~when ~~~t={1\over6}\pi$$

Question

order: Find the gradient of the normal to the curve, at the specified value of t. 
$$x=\cos^3t~~~~y=\sin^3t~~~~~~when ~~~t={1\over6}\pi$$

anonymous · Answer

@dumbcow

anonymous · Answer

x'= -3cos^2tsint???
y'= 3sin^2tcost???

dumbcow · Answer

yes :)

anonymous · Answer

$$3\sin^2tcost \over -3\cos^2tsint$$

anonymous · Answer

$$-sint \over cost$$?

dumbcow · Answer

good

anonymous · Answer

-tant?

dumbcow · Answer

yes either way would work since you still have to evaluate at t=pi/6

anonymous · Answer

Ok. let me evaluate. How can you do this using pi ? I always convert to decimals...?

dumbcow · Answer

pi/6 converted to degrees is 180/6 or 30

anonymous · Answer

OK
-tan30= -0.57735
m(-0.57735)=-1
m=1.73205

How do I convert this into sqrt?

dumbcow · Answer

looks good
to leave it in radicals
-tan(30) = -1/sqrt3

m = sqrt3

anonymous · Answer

Is there some sort of formula for radicals?

dumbcow · Answer

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 · Answer

Hm Ok. Thank you! In exams, do they regard radicals over decimals? Or do you not know?

dumbcow · Answer

depends on the teacher or policies of school
radical is considered more exact while decimal are approximate answers

anonymous · Answer

Hm Ok. It's external examinations.

anonymous · Answer

is this trigonometry?  @dum

anonymous · Answer

@dumbcow

anonymous · Answer

Trig, yes and parametric equations

dumbcow · Answer

yeah ^^

anonymous · Answer

ooh, more advance,.

anonymous · Answer

Thanks for helping!