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

Question

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

Gregory Ferguson · Accepted Answer

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&#039;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&#039;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

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

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