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  • If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the parameter, Find dy/dx x = a cos theta +theta sin theta y = a sin theta - theta cos theta

Q10  If x and y are connected parametrically by the equations given in Exercises 1 to 10, without eliminating the  parameter, Find dy/dx x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )

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best_answer

Given equations are
x = a ( \cos \theta + \theta \sin \theta ) , y = a ( \sin \theta - \theta \cos \theta )
Now, differentiate both w.r.t  \theta
We get,
\frac{dx}{d\theta}=\frac{d(a(\cos \theta+ \theta\sin \theta))}{d\theta}= a(-\sin \theta+\sin \theta+ \theta\cos \theta)= a \theta\cos \theta
Similarly,
\frac{dy}{d\theta}=\frac{d(a(\sin \theta- \theta\cos \theta))}{d\theta}= a(\cos \theta-\cos \theta+ \theta\sin \theta) = a \theta\sin \theta
Now, \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}= \frac{a \theta\sin \theta}{a \theta\cos \theta} = \tan \theta
Therefore, the answer is \frac{dy}{dx}= \tan \theta
 

Posted by

Gautam harsolia

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