Question

If \[{{x = 2cos\theta - cos2\theta }}\] and \[{{y = 2sin\theta - sin2\theta }}\] find \[\dfrac{{{{{d}}^{{2}}}{{y}}}}{{{{d}}{{{x}}^{{2}}}}}\] at \[{{\theta = }}\dfrac{{{\pi }}}{{{2}}}\]

Answer 1

Hint: Here In this question we need to find the second order derivative at the given particular degree
We are going solve the using the values which they have given in the question
For Solving this question using the differentiation formula
We need the find these by applying the values on the differentiation formula we will get the values which will help to solve
Again, we need to differentiate them
we will get the second order derivative and applying the mentioned degree on them
Finally, we will have the required answer

Formula used: The formulas which we are using in the equations are,
\[\dfrac{{{d}}}{{{{dx}}}}{{(sinx) = cosx}}\]
\[\dfrac{{{d}}}{{{{dx}}}}{{(cosx) = - sinx}}\]
\[{{sinA - sinB = 2 cos}}\dfrac{{{{A + B}}}}{{{2}}}{{.sin}}\dfrac{{{{A - B}}}}{{{2}}}\]
\[{{cosA - cosB = 2 sin}}\dfrac{{{{A + B}}}}{{{2}}}{{.sin}}\dfrac{{{{A - B}}}}{{{2}}}\]

Complete step-by-step answer:
According to this question we have that
\[{{x = 2cos\theta - cos2\theta }}\] …………………………..\[{{(1)}}\]
\[{{y = 2sin\theta - sin2\theta }}\] ……………………………\[{{(2)}}\]
By Differentiate equation \[{{(1)}}\] we will get,
$\Rightarrow$\[\dfrac{{{{dx}}}}{{{{d\theta }}}}\]\[{{ = - 2 sin\theta + 2 sin2\theta }}\]
By Differentiate equation \[{{(2)}}\] we will get,
$\Rightarrow$\[\dfrac{{{{dy}}}}{{{{d\theta }}}}\]\[{{ = 2 cos\theta - 2 cos2\theta }}\]
Here, we are dividing \[\dfrac{{{{dx}}}}{{{{d\theta }}}}\] and $\Rightarrow$\[\dfrac{{{{dy}}}}{{{{d\theta }}}}\], we will have that
$\Rightarrow$\[\dfrac{{{{dy}}}}{{{{dx}}}}\]\[{{ = }}\dfrac{{{{2 cos\theta - 2 cos2\theta }}}}{{{{ - 2 sin\theta + 2 sin2\theta }}}}\]
As we know the common term on the above values is \[{{2}}\],
We are cancelling out them,
$\Rightarrow$\[\dfrac{{{{dy}}}}{{{{dx}}}}\]\[{{ = }}\dfrac{{{{cos\theta - cos2\theta }}}}{{{{ sin2\theta - sin\theta }}}}\]
Here, we are applying \[{{cosA - cosB}}\] formula and \[{{sinA - sinB}}\] formula, we will get
$\Rightarrow$\[\dfrac{{{{dy}}}}{{{{dx}}}}\]\[{{ = }}\dfrac{{{{2 sin}}\left( {\dfrac{{{{2\theta + \theta }}}}{{{2}}}} \right){{sin}}\left( {\dfrac{{{{2\theta - \theta }}}}{{{2}}}} \right){{ }}}}{{{{2 cos}}\left( {\dfrac{{{{2\theta + \theta }}}}{{{2}}}} \right){{sin}}\left( {\dfrac{{{{2\theta - \theta }}}}{{{2}}}} \right){{ }}}}\]
Cancelling out the like terms on the above, we have,
$\Rightarrow$\[\dfrac{{{{dy}}}}{{{{dx}}}}\]\[{{ = }}\dfrac{{{{ sin}}\left( {\dfrac{{{{2\theta + \theta }}}}{{{2}}}} \right){{ }}}}{{{{ cos}}\left( {\dfrac{{{{2\theta + \theta }}}}{{{2}}}} \right){{ }}}}\]
Let us consider, \[\dfrac{{{{sin\theta }}}}{{{{cos\theta }}}}{{ = tan\theta }}\]
$\Rightarrow$\[\dfrac{{{{dy}}}}{{{{dx}}}}\]\[{{ = tan}}\left( {\dfrac{{{{3\theta }}}}{{{2}}}} \right)\]
Again, we differentiate \[\dfrac{{{{dy}}}}{{{{dx}}}}\], we get,
$\Rightarrow$\[\dfrac{{\dfrac{{{d}}}{{{{d\theta }}}}\left( {\dfrac{{{{dy}}}}{{{{dx}}}}} \right)}}{{\dfrac{{{{dx}}}}{{{{d\theta }}}}}}\] \[{{ = }}\dfrac{{\dfrac{{{3}}}{{{2}}}{{se}}{{{c}}^{{2}}}\left( {\dfrac{{{{3\theta }}}}{{{2}}}} \right)}}{{{{2 sin2\theta - 2sin\theta }}}}\]\[ = \dfrac{{{{{d}}^{{2}}}{{y}}}}{{{{d}}{{{x}}^{{2}}}}}\]
$\Rightarrow$\[\dfrac{{{{{d}}^{{2}}}{{y}}}}{{{{d}}{{{x}}^{{2}}}}}{{ = }}\dfrac{{\dfrac{{{3}}}{{{2}}}{{se}}{{{c}}^{{2}}}\left( {\dfrac{{{{3\theta }}}}{{{2}}}} \right)}}{{{{2 sin2\theta - 2sin\theta }}}}\]
At \[{{\theta = }}\dfrac{{{\pi }}}{{{2}}}\],
By Applying \[{{\theta }}\] value, we get
$\Rightarrow$\[\dfrac{{{{{d}}^{{2}}}{{y}}}}{{{{d}}{{{x}}^{{2}}}}}{{ = }}\]\[{{ }}\dfrac{{{3}}}{{{4}}}\left( {\dfrac{{{2}}}{{{{ - 1}}}}} \right)\]
By cancelling each other, we get,
$\Rightarrow$\[\dfrac{{{{{d}}^{{2}}}{{y}}}}{{{{d}}{{{x}}^{{2}}}}}{{ = }}\]\[{{ - }}\dfrac{{{3}}}{2}\]

Therefore, the second order derivative at the given point \[\dfrac{{{\pi }}}{{{2}}}\] is \[{{ - }}\dfrac{{{3}}}{2}\]

Note: In this question we have
\[{{x = 2cos\theta - cos2\theta }}\] and \[{{y = 2sin\theta - sin2\theta }}\]
While we differentiate them, we get,
$\Rightarrow$\[\dfrac{{{{dx}}}}{{{{d\theta }}}}\]\[{{ = - 2 sin\theta + 2 sin2\theta }}\]
$\Rightarrow$\[{{y = 2sin\theta - sin2\theta }}\]
$\Rightarrow$\[\dfrac{{{{dy}}}}{{{{d\theta }}}}\]\[{{ = 2 cos\theta - 2 cos2\theta }}\]
Here we are finding \[\dfrac{{{{dy}}}}{{{{dx}}}}\],
By chain rule,
$\Rightarrow$\[\dfrac{{{{dy}}}}{{{{dx}}}}{{ = }}\dfrac{{{{dy}}}}{{{{d\theta }}}}{{ \times }}\dfrac{{{{d\theta }}}}{{{{dx}}}}\]
$\Rightarrow$\[\dfrac{{{{dy}}}}{{{{dx}}}}\]\[ = \dfrac{{{{ 2cos\theta - 2cos2\theta }}}}{{{{ - 2 sin\theta + 2sin2\theta }}}}\]
Here, we are differentiating them again we get
$\Rightarrow$\[\dfrac{{{{{d}}^{{2}}}{{y}}}}{{{{d}}{{{x}}^{{2}}}}}{{ = }}\dfrac{{{d}}}{{{{dt}}}}\left( {\dfrac{{{{dy}}}}{{{{dx}}}}} \right)\dfrac{{{{dt}}}}{{{{dx}}}}\]
$\Rightarrow$\[\dfrac{{{{{d}}^{{2}}}{{y}}}}{{{{d}}{{{x}}^{{2}}}}}{{ = }}\dfrac{{{d}}}{{{{dt}}}}\]\[\left( {\dfrac{{{{ cos\theta - cos2\theta }}}}{{{{ sin2\theta - sin\theta }}}}} \right) \times \dfrac{1}{{{{ - 2 sin\theta + 2 sin2\theta }}}}\]
While multiplying we get
$\Rightarrow$\[\dfrac{{{{{d}}^{{2}}}{{y}}}}{{{{d}}{{{x}}^{{2}}}}}{{ = }}\dfrac{{{{sin2\theta - sin\theta (2sin2\theta - sin\theta ) - (cos\theta - cos2\theta )(2 cos2\theta - cos\theta )}}}}{{{{{{(sin2\theta - sin\theta )}}}^{{2}}}}}{{ \times }}\dfrac{{{1}}}{{{{ - 2 sin\theta + 2 sin2\theta }}}}\]
At \[{{\theta = }}\dfrac{{{\pi }}}{{{2}}}\],
By applying \[{{\theta }}\] value, we get
\[ = \dfrac{{( - 1)( - 1) - (1)( - 2)}}{{{{( - 1)}^2}}} \times \dfrac{1}{{( - 2)}}\]
By multiplying
$\Rightarrow$\[\dfrac{{{{{d}}^{{2}}}{{y}}}}{{{{d}}{{{x}}^{{2}}}}}\]\[{{ = - }}\dfrac{{{3}}}{{{2}}}\]
Therefore, the second order derivative at the given point \[\dfrac{{{\pi }}}{{{2}}}\] is \[{{ - }}\dfrac{{{3}}}{2}\]