Solve for dy/dx?

r=6\cos\theta, \theta=(3\pi)/2

I know to use the formula (r'\sin\theta+r\cos\theta)/(r'\cos\theta-r\sin\theta)

1 Answer
May 25, 2018

Answer: dy/dx|_(theta=(3pi)/2) is undefined

Explanation:

Consider that (dy)/(dx)=(r'sintheta+rcostheta)/(r'costheta-rsintheta) where r'=(dr)/(d theta)

Since we are given that r=6cos(theta), we can take the derivative of r with respect to theta to find r':
r'=-6sin(theta)

Plugging the equation for r and r' into the equation for dy/dx, we have:
dy/dx=((-6sintheta)(sintheta)+(6costheta)(costheta))/((-6sintheta)(costheta)-(6costheta)(sintheta))
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If we are solving for dy/dx in general, we can continue to simply this expression:
dy/dx=(6(cos^2theta-sin^2theta))/(6(-2sinthetacostheta)

Consider the double-angle formulas: sin(2theta)=2sinthetacostheta and cos(2theta)=cos^2theta-sin^2theta

Applying these formulas we have:
dy/dx=-cos(2theta)/sin(2theta)=-cot(2theta)
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However, since we are given an angle (theta=(3pi)/2), if we are actually asked to find dy/dx|_(theta=(3pi)/2), then we can simply use the original plugged in expression:
dy/dx=((-6sintheta)(sintheta)+(6costheta)(costheta))/((-6sintheta)(costheta)-(6costheta)(sintheta))

Noting that cos((3pi)/2)=0, we can eliminate all terms that have a costheta:

dy/dx|_(theta=(3pi)/2)=((-6sintheta)(sintheta)+0)/(0)

Notice that we have division by 0 when we do this substitution, this indicates that dx/(d theta)=0, so we have a vertical tangent and dy/dx|_(theta=(3pi)/2) is undefined.