What is the slope of the tangent line of r=-2sin(3theta)-12cos(theta/2) at theta=(-pi)/3?

Teresa Manning

Teresa Manning

Answered question

2023-03-14

What is the slope of the tangent line of r = - 2 sin ( 3 θ ) - 12 cos ( θ 2 ) at θ = - π 3 ?

Answer & Explanation

Gregory Ferguson

Gregory Ferguson

Beginner2023-03-15Added 3 answers

The slope of the tangent line of r at θ = - π 3 , is the same as the derivative of the function for that exact x-value. Therefore, we need to differentiate both sides of the function, so we get an expression for d d θ r.
d d θ [ r ] = d d θ [ - 2 sin ( 3 θ ) - 12 cos ( θ 2 ) ]
d d θ [ r ] = d d θ [ - 2 sin ( 3 θ ) ] - d d θ [ - 12 cos ( θ 2 ) ]
From this point, we need to know how to differentiate sin θ and cos θ :
d d θ sin θ = cos θ
d d θ cos θ = - sin θ
We need to know how the chain rule works as well. In this case, we have to substitute 3 θ with u and θ 2 with v. When we then differentiate, we have to also differentiate our substitute.
We then have:
d d θ [ r ] = d d θ [ - 2 sin ( u ) ] - d d θ [ 12 cos ( v ) ]
d d θ [ r ] = - 2 cos ( u ) d d θ [ u ] - ( - 12 sin ( v ) d d θ [ v ] )
Let's now substitute u and v back to our original functions.
d d θ [ r ] = - 2 cos ( 3 θ ) d d θ [ 3 θ ] - ( - 12 sin ( θ 2 ) d d θ [ θ 2 ] )
d d θ [ r ] = - 2 cos ( 3 θ ) 3 + 12 sin ( θ 2 ) 1 2 = - 6 cos ( 3 θ ) + 6 sin ( θ 2 )
Now we have an expression for d d θ r ( θ ) , where we can put in any values we want, and get the slope for whatever θ -value we want. Thus, let's put in θ = - π 3 :
d d θ [ r ( - π 3 ) ]
= - 6 cos ( - π ) + 6 sin ( - π 6 )
cos ( - π ) = - 1 and sin ( - π 6 ) = - 1 2
This gives us that the slope, for θ = ( - π 3 ) , is:
- 6 ( - 1 ) + 6 ( - 1 2 ) = 6 - 3 = 3

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