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Luba_88 [7]
3 years ago
6

More Calculus! (I'm so sorry)

Mathematics
1 answer:
Olenka [21]3 years ago
6 0
Recall that converting from Cartesian to polar coordinates involves the identities

\begin{cases}y(r,\phi)=r\sin\phi\\x(r,\phi)=r\cos\phi\end{cases}

As a function in polar coordinates, r depends on \phi, so you can write r=r(\phi).

Differentiating the identities with respect to \phi gives

\begin{cases}\dfrac{\mathrm dy}{\mathrm d\phi}=\dfrac{\mathrm dr}{\mathrm d\phi}\sin\phi+r\cos\phi\\\\\dfrac{\mathrm dx}{\mathrm d\phi}=\dfrac{\mathrm dr}{\mathrm d\phi}\cos\phi-r\sin\phi\end{cases}

The slope of the tangent line to r(\phi) is given by

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm d\phi}}{\frac{\mathrm dx}{\mathrm d\phi}}=\dfrac{\frac{\mathrm dr}{\mathrm d\phi}\sin\phi+r\cos\phi}{\frac{\mathrm dr}{\mathrm d\phi}\cos\phi-r\sin\phi}

Given r(\phi)=3\cos\phi, you have \dfrac{\mathrm dr}{\mathrm d\phi}=-3\sin\phi. So the tangent line to r(\phi) has a slope of

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{-3\sin^2\phi+3\cos^2\phi}{-3\sin\phi\cos\phi-3\cos\phi\sin\phi}=\dfrac{3\cos2\phi}{-3\sin2\phi}=-\cot2\phi

When \phi=120^\circ=\dfrac{2\pi}3\text{ rad}, the tangent line has slope

\dfrac{\mathrm dy}{\mathrm dx}=-\cot\dfrac{4\pi}3=-\dfrac1{\sqrt3}

This line is tangent to the point (r,\phi)=\left(-\dfrac32,\dfrac{2\pi}3\right) which in Cartesian coordinates is equivalent to (x,y)=\left(\dfrac34,-\dfrac{3\sqrt3}4\right), so the equation of the tangent line is

y+\dfrac{3\sqrt3}4=-\dfrac1{\sqrt3}\left(x-\dfrac34\right)

In polar coordinates, this line has equation

r\sin\phi+\dfrac{3\sqrt3}4=-\dfrac1{\sqrt3}\left(r\cos\phi-\dfrac34\right)
\implies r=-\dfrac{3\sqrt3}{2\sqrt3\cos\phi+6\sin\phi}

The tangent line passes through the y-axis when x=0, so the y-intercept is \left(0,-\dfrac{\sqrt3}2\right).

The vector from this point to the point of tangency on r(\phi) is given by the difference of the vector from the origin to the y-intercept (which I'll denote \mathbf a) and the vector from the origin to the point of tangency (denoted by \mathbf b). In the attached graphic, this corresponds to the green arrow.

\mathbf b-\mathbf a=\left(\dfrac34,-\dfrac{3\sqrt3}4\right)-\left(0,-\dfrac{\sqrt3}2\right)=\left(\dfrac34,-\dfrac{\sqrt3}4\right)

The angle between this vector and the vector pointing to the point of tangency is what you're looking for. This is given by

\mathbf b\cdot(\mathbf b-\mathbf a)=\|\mathbf b\|\|\mathbf b-\mathbf a\|\cos\theta
\dfrac98=\dfrac{3\sqrt3}4\cos\theta
\implies\theta=\dfrac\pi6\text{ rad}=30^\circ

The second problem is just a matter of computing the second derivative of \phi with respect to t and plugging in t=2.

\phi(t)=2t^3-6t
\phi'(t)=6t^2-6
\phi''(t)=12t
\implies\phi''(2)=24

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