- The two component-rate formulas are already supplied; this is a direct substitution task. - Use the exact sine and cosine values at \(\theta=0\). - Keep the x-component rate and y-component rate separate.

1. Substitute \(\theta=0\), \(r=3\), and \(r'=0\). 2. Then \(\frac{dx}{d\theta}=0\cdot1-3\cdot0=0\). 3. Also, \(\frac{dy}{d\theta}=0\cdot0+3\cdot1=3\).

\(\frac{dx}{d\theta}=0\), \(\frac{dy}{d\theta}=3\)

- Both supplied rates use the same angular parameter. - The requested slope measures change in \(y\) per unit change in \(x\). - Check the sign before simplifying the ratio.

1. The Cartesian slope compares the y-rate with the x-rate for the same parameter \(\theta\). 2. Thus, \(\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\frac{6}{-2}=-3\).

\(-3\)

- Locate the ray for a positive radius at the stated angle. - What does a negative radius do to that direction?

1. The direction \(\theta=\frac{\pi}{2}\) points upward along the positive y-axis. 2. A negative radius sends the point the same distance in the opposite direction, placing P at \((0,-2)\).

\((-2,\frac{\pi}{2})\).

- Evaluate the radius at the stated angle. - What location does a radius of zero represent?

1. At zero angle, \(r=2\sin0=0\). 2. A zero polar radius places the point at the pole.

At the pole.

- Increasing \(\theta\) means counterclockwise travel around the circle. - A tangent to a circle is perpendicular to its radius at the point.

1. At P the radius points to the right. 2. Increasing the angle moves the point upward along the circle, so its tangent is vertical.

Vertical.

- Changing radius at a fixed angle moves along a straight ray. - Which marked point is farther from the pole?

1. P and Q lie on the same ray from the origin. 2. Q is farther from the origin, so moving from P to Q makes the radius larger.

P to Q.

- Differentiate the radius as a function of the angle. - Is changing the radius the same as changing the point's position?

1. The radius is constant, so \(\frac{dr}{d\theta}=0\). 2. The point still moves around a circle when the angle changes.

\(\frac{dr}{d\theta}=0\); no, the point moves around the circle.

- Think of the polar curve as a parametrized point whose rectangular coordinates both depend on \(\theta\). - When differentiating those coordinates, remember that the radius is changing as well as the trigonometric factors. - Use the two resulting rectangular rates to determine the tangent slope at the requested angle.

1. Treat the polar curve parametrically: \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\), with \(r'=-\sin(\theta)\). 2. Then \(\frac{dx}{d\theta}=r'\cos(\theta)-r\sin(\theta)=-2(1+\cos(\theta))\sin(\theta)\). 3. Also, \(\frac{dy}{d\theta}=r'\sin(\theta)+r\cos(\theta)=2\cos(\theta)+\cos(2\theta)\). 4. At \(\theta=\frac{\pi}{2}\), \(\frac{dx}{d\theta}=-2\) and \(\frac{dy}{d\theta}=-1\), so \(\frac{dy}{dx}=\frac{1}{2}\).

\(\frac{dy}{dx}=\frac{1}{2}\)

- Express the polar point in Cartesian coordinates. - Both coordinates change when the angle changes. - Simplify the trigonometric derivatives before substituting the angle.

1. The Cartesian coordinates are \(x=3\sin\theta\cos\theta\) and \(y=3\sin^{2}\theta\). 2. Differentiate: \(\frac{dx}{d\theta}=3\cos(2\theta)\) and \(\frac{dy}{d\theta}=3\sin(2\theta)\). 3. At \(\theta=\frac{\pi}{6}\), the rates are \(\frac{3}{2}\) and \(\frac{3\sqrt{3}}{2}\). 4. Their ratio gives \(\frac{dy}{dx}=\sqrt{3}\).

\(\frac{dx}{d\theta}=\frac{3}{2}\). \(\frac{dy}{d\theta}=\frac{3\sqrt{3}}{2}\). \(\frac{dy}{dx}=\sqrt{3}\).

- Determine how both rectangular coordinates change as \(\theta\) changes. - Keep the radial derivative and the trigonometric derivatives separate until the component rates are correct. - Before forming a finite slope, check the horizontal component rate at the given angle.

1. Use \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\), with \(r'=-2\sin(\theta)\). 2. Then \(\frac{dx}{d\theta}=-(4\cos(\theta)+1)\sin(\theta)\) and \(\frac{dy}{d\theta}=\cos(\theta)+2\cos(2\theta)\). 3. At \(\theta=\frac{\pi}{2}\), these derivatives are \(-1\) and \(-2\), so \(\frac{dy}{dx}=2\).

\(\frac{dy}{dx}=2\)

- The radial function itself has a composite trigonometric argument, so differentiate it carefully. - At the target angle, evaluate the radial information needed for the rectangular component rates. - Use those component rates to obtain the tangent direction without treating \(dr/d\theta\) as the slope.

1. Use \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\), with \(r'=4\cos(2\theta)\). 2. The derivatives are \(\frac{dx}{d\theta}=r'\cos(\theta)-r\sin(\theta)\) and \(\frac{dy}{d\theta}=r'\sin(\theta)+r\cos(\theta)\). 3. At \(\theta=\frac{\pi}{6}\), \(r=\sqrt{3}\) and \(r'=2\), so \(\frac{dx}{d\theta}=\frac{\sqrt{3}}{2}\) and \(\frac{dy}{d\theta}=\frac{5}{2}\). 4. Therefore, \(\frac{dy}{dx}=\frac{5/2}{\sqrt{3}/2}=\frac{5\sqrt{3}}{3}\).

\(\frac{dy}{dx}=\frac{5\sqrt{3}}{3}\)

- The rectangular position changes for two reasons: the radius changes and the angle changes. - Differentiate each rectangular coordinate with both sources of change included. - Evaluate the resulting direction rates at \(\theta=0\) before interpreting the slope.

1. Use \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\), with \(r'=e^{\theta}\). 2. Then \(\frac{dx}{d\theta}=e^{\theta}(\cos(\theta)-\sin(\theta))\) and \(\frac{dy}{d\theta}=e^{\theta}(\sin(\theta)+\cos(\theta))\). 3. At \(\theta=0\), both derivatives equal \(1\), so \(\frac{dy}{dx}=1\).

\(\frac{dy}{dx}=1\)

- Differentiate the radial function carefully before combining radial and angular change. - Simplify the rectangular component rates enough that the special angle is easy to evaluate. - Keep track of both signs when converting the two component rates into a tangent slope.

1. Use \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\), with \(r'=\frac{4\sin(\theta)}{(1+\cos(\theta))^{2}}\). 2. The component derivatives simplify to \(\frac{dx}{d\theta}=-\frac{4\sin(\theta)}{(1+\cos(\theta))^{2}}\) and \(\frac{dy}{d\theta}=\frac{4}{1+\cos(\theta)}\). 3. At \(\theta=\frac{\pi}{2}\), these are \(-4\) and \(4\), so \(\frac{dy}{dx}=-1\).

\(\frac{dy}{dx}=-1\)

- The table gives local radial information rather than a complete formula for \(r(\theta)\); decide what local rectangular information it determines. - Use the special-angle trigonometric values when converting radial and angular change into x- and y-rates. - Check whether the horizontal rate permits a finite slope before dividing.

1. Use \(\frac{dx}{d\theta}=r'\cos(\theta)-r\sin(\theta)\) and \(\frac{dy}{d\theta}=r'\sin(\theta)+r\cos(\theta)\). 2. At \(\theta=0\), the data give \(\frac{dx}{d\theta}=1\) and \(\frac{dy}{d\theta}=2\). 3. Since the horizontal rate is nonzero, the tangent slope is \(\frac{dy}{dx}=2\).

\(\frac{dy}{dx}=2\)

- Use both pieces of local radial data; neither one alone determines the rectangular tangent. - Evaluate how the point's x- and y-coordinates are changing at \(\theta=\frac{\pi}{2}\). - Preserve the signs of those rates when deciding whether the tangent is finite or vertical.

1. Use \(\frac{dx}{d\theta}=r'\cos(\theta)-r\sin(\theta)\) and \(\frac{dy}{d\theta}=r'\sin(\theta)+r\cos(\theta)\). 2. At \(\theta=\frac{\pi}{2}\), the data give \(\frac{dx}{d\theta}=-1\) and \(\frac{dy}{d\theta}=-2\). 3. Therefore, \(\frac{dy}{dx}=\frac{-2}{-1}=2\).

\(\frac{dy}{dx}=2\)

- Translate the local radial data into the two rectangular component rates at \(\theta=\pi\). - A tangent's orientation depends on the relationship between those two rates, not on \(dr/d\theta\) alone. - Decide whether a finite quotient is possible after evaluating the component rates.

1. Use the rectangular component derivative formulas for a polar curve. 2. At \(\theta=\pi\), the data give \(\frac{dx}{d\theta}=0\) and \(\frac{dy}{d\theta}=-3\). 3. Because the horizontal rate is zero while the vertical rate is nonzero, the tangent is vertical.

The tangent is vertical.

- Use the indicated angle to determine radial change and both Cartesian rates. - Compare the two coordinate rates before deciding the tangent direction. - Check whether the computed tangent direction matches the dashed segment from the origin to P.

1. At \(\theta=0\), \(r=2\) and \(\frac{dr}{d\theta}=\cos(0)=1\). 2. From \(x=r\cos\theta\), \(\frac{dx}{d\theta}=r'\cos\theta-r\sin\theta=1\). 3. From \(y=r\sin\theta\), \(\frac{dy}{d\theta}=r'\sin\theta+r\cos\theta=2\). 4. The tangent slope is \(\frac{dy}{dx}=\frac{2}{1}=2\). The ray from the origin through \(P=(2,0)\) has slope \(0\), so the tangent is not radial.

\(\frac{dr}{d\theta}=1\), \(\frac{dx}{d\theta}=1\), \(\frac{dy}{d\theta}=2\), \(\frac{dy}{dx}=2\). The tangent is not along the radial ray.

- Use the rectangular coordinate rates as a directed tangent vector. - Preserve the orientation associated with increasing angle. - Normalize the nonzero vector after evaluating it.

1. The radial derivative is \(r'=-2\sin(2\theta)\), so at \(\theta=0\), \(r=2\) and \(r'=0\). 2. The rectangular coordinate rates are \(\frac{dx}{d\theta}=0\) and \(\frac{dy}{d\theta}=2\). 3. The tangent vector for increasing \(\theta\) is \(\langle0,2\rangle\). 4. Dividing by its magnitude gives the unit tangent vector \(\langle0,1\rangle\).

\(\langle0,1\rangle\)

- The derivative of a polar radius is not the same thing as Cartesian tangent slope. - Find both Cartesian coordinate rates at the stated angle. - Which rate is zero for a vertical tangent?

1. At \(\theta=0\), \(r=1\) and \(\frac{dr}{d\theta}=0\), so the point is \((1,0)\). 2. The Cartesian coordinate rates are \(\frac{dx}{d\theta}=r'\cos\theta-r\sin\theta=0\) and \(\frac{dy}{d\theta}=r'\sin\theta+r\cos\theta=1\). 3. The tangent is vertical, with equation \(x=1\), so the claim is false.

The claim is false. The tangent at \(\theta=0\) is vertical: \(x=1\).

- Convert the polar circle to coordinate functions of the angle. - Use the ratio of the vertical and horizontal rates.

1. Since the radius is constant, \(\frac{dx}{d\theta}=-2\sin\theta\) and \(\frac{dy}{d\theta}=2\cos\theta\). 2. Their quotient at \(\frac{\pi}{4}\) is \(-\frac{\cos(\pi/4)}{\sin(\pi/4)}=-1\).

\(-1\).

- Both the radius and the angle change here. - Apply the product rule to the Cartesian coordinate expressions. - Evaluate the two rates at zero before taking their quotient.

1. Here \(r(0)=1\) and \(r'(0)=1\). 2. From \(x=r\cos\theta\) and \(y=r\sin\theta\), we get \(x'=r'\cos\theta-r\sin\theta\) and \(y'=r'\sin\theta+r\cos\theta\). 3. At zero the rates are both \(1\), giving slope \(1\).

\(1\).

- Use the locations of P and Q to identify their polar angles. - At each angle, compute the radius and its rate of change. - A vertical tangent needs a zero horizontal rate and a nonzero vertical rate.

1. P lies at \(\theta=0\). There \(r=\frac{3}{2}\), \(r'=0\), and the Cartesian rates are \(\frac{dx}{d\theta}=0\), \(\frac{dy}{d\theta}=\frac{3}{2}\). 2. Q lies at \(\theta=\frac{\pi}{2}\). There \(r=1\), \(r'=-\frac{1}{2}\), giving \(\frac{dx}{d\theta}=-1\) and \(\frac{dy}{d\theta}=-\frac{1}{2}\). 3. Only P has zero horizontal rate and nonzero vertical rate, so only P has a vertical tangent.

P. At P the rates are \((0,\frac{3}{2})\); at Q they are \((-1,-\frac{1}{2})\).

- A negative radius places the point opposite the positive-angle ray. - Write its Cartesian coordinates before checking the motion direction.

1. The Cartesian position is \((x,y)=(-\cos\theta,-\sin\theta)\). 2. Differentiate: \((x',y')=(\sin\theta,-\cos\theta)\). 3. At zero the direction vector is \((0,-1)\), pointing down.

Down.

- Use the polar-to-Cartesian rate formulas at the special angle zero. - Work backward from the given tangent slope rather than choosing a full formula for \(r(\theta)\).

1. At zero angle, the Cartesian coordinate rates are \(x'=r'\) and \(y'=r\). 2. The slope condition is \(\frac{r}{r'}=3\), or \(\frac{2}{r'}=3\). 3. Thus \(r'=\frac{2}{3}\), which is nonzero as needed.

\(\frac{dr}{d\theta}=\frac{2}{3}\) at \(\theta=0\).

- Keep the negative sign on the radius when differentiating the Cartesian coordinates. - At a right angle, the sine and cosine values simplify the rates.

1. The horizontal rate is \(x'=r'\cos\theta-r\sin\theta=0-(-1)=1\). 2. The vertical rate is \(y'=r'\sin\theta+r\cos\theta=2\). 3. The slope is \(\frac{dy}{dx}=\frac{2}{1}=2\).

\(2\).

- Find both the Cartesian point and the coordinate rates. - Do not treat the polar radius as the Cartesian tangent slope. - Use the computed point and direction to form a line equation.

1. At \(\theta=\frac{\pi}{4}\), \(r=\sqrt{2}\) and the point is \((1,1)\). 2. Since \(\frac{dr}{d\theta}=-2\sin\theta\), the Cartesian rates are \(\frac{dx}{d\theta}=-2\) and \(\frac{dy}{d\theta}=0\). 3. The tangent slope is \(0\), so the tangent line is \(y=1\).

Point: \((1,1)\). \(\frac{dx}{d\theta}=-2\), \(\frac{dy}{d\theta}=0\), \(\frac{dy}{dx}=0\). Tangent line: \(y=1\).

- Locate the polar point in rectangular coordinates and determine the tangent direction there. - The requested line is perpendicular to the tangent, so compare their allowable slopes or directions. - Check that your final line passes through the point on the curve.

1. At \(\theta=0\), \(r=1\), so the rectangular point is \((1,0)\). 2. With \(r'=\cos(\theta)\), the component derivatives at \(\theta=0\) are \(\frac{dx}{d\theta}=1\) and \(\frac{dy}{d\theta}=1\), giving tangent slope \(1\). 3. The normal slope is the negative reciprocal, \(-1\). 4. The normal line is \(y=-(x-1)\).

\(y=-(x-1)\)

- Find the point and tangent direction separately. - Account for the changing radius when finding the coordinate rates. - Point-slope form keeps the tangent line exact.

1. At \(\theta=\frac{\pi}{3}\), the radius is \(\frac{3\sqrt{3}}{2}\), giving \(\left(\frac{3\sqrt{3}}{4},\frac{9}{4}\right)\). 2. Since \(\frac{dr}{d\theta}=3\cos\theta\), the Cartesian coordinate rates are \(\frac{dx}{d\theta}=-\frac{3}{2}\) and \(\frac{dy}{d\theta}=\frac{3\sqrt{3}}{2}\). 3. The tangent slope is \(-\sqrt{3}\). 4. The line is \(y-\frac{9}{4}=-\sqrt{3}\left(x-\frac{3\sqrt{3}}{4}\right)\).

Point: \(\left(\frac{3\sqrt{3}}{4},\frac{9}{4}\right)\). \(\frac{dx}{d\theta}=-\frac{3}{2}\), \(\frac{dy}{d\theta}=\frac{3\sqrt{3}}{2}\), \(\frac{dy}{dx}=-\sqrt{3}\). Tangent: \(y-\frac{9}{4}=-\sqrt{3}\left(x-\frac{3\sqrt{3}}{4}\right)\).

- Determine the rectangular point and the tangent direction associated with the given polar angle. - Use perpendicularity to relate the normal direction to the tangent direction. - Verify the resulting normal line passes through the correct point.

1. At \(\theta=\frac{\pi}{2}\), \(r=2\), so the rectangular point is \((0,2)\). 2. With \(r'=-2\sin(\theta)\), the component derivatives are \(\frac{dx}{d\theta}=-2\) and \(\frac{dy}{d\theta}=-2\), so the tangent slope is \(1\). 3. The normal slope is \(-1\). 4. Therefore, the normal line is \(y-2=-x\).

\(y-2=-x\)

- Evaluate \(r\) and \(r'\) at the specified angle before forming the slope. - Use the rectangular component derivatives rather than the radial derivative alone. - Set the resulting slope expression equal to the given value and verify its denominator is nonzero.

1. At \(\theta=0\), \(r=a\) and \(r'=1\). 2. The rectangular component derivatives are \(\frac{dx}{d\theta}=r'\cos(\theta)-r\sin(\theta)=1\) and \(\frac{dy}{d\theta}=r'\sin(\theta)+r\cos(\theta)=a\). 3. Thus, the tangent slope is \(a\). Setting \(a=2\) gives the required slope, and \(dx/d\theta\ne0\).

\(a=2\)

- The prescribed rectangular slope places a condition on how the polar point is moving at \(\theta=0\). - Express the needed rectangular component rates in terms of the unknown parameter \(a\). - After solving for \(a\), confirm that the tangent slope is actually defined.

1. At \(\theta=0\), \(r=1\) and \(r'=a\). 2. The component derivatives are \(\frac{dx}{d\theta}=a\) and \(\frac{dy}{d\theta}=1\), so the tangent slope is \(\frac{1}{a}\). 3. Setting \(\frac{1}{a}=\frac{1}{2}\) gives \(a=2\), for which \(dx/d\theta\ne0\).

\(a=2\)

- Evaluate the radial behavior at the stated angle in terms of \(a\). - Translate the prescribed tangent slope into a relationship between the two rectangular component rates. - Check the horizontal rate after finding the parameter so the finite slope is valid.

1. At \(\theta=\frac{\pi}{2}\), \(r=1\) and \(r'=-a\). 2. The component derivatives are \(\frac{dx}{d\theta}=-1\) and \(\frac{dy}{d\theta}=-a\), so the tangent slope is \(a\). 3. Setting \(a=-\frac{1}{2}\) gives the required slope, and the horizontal rate remains nonzero.

\(a=-\frac{1}{2}\)

- At \(\theta=0\), express both the radius and its rate of change in terms of \(a\). - The known tangent slope constrains the ratio of the rectangular component rates. - Verify that your parameter value does not make that ratio undefined.

1. At \(\theta=0\), \(r=1+a\) and \(r'=a\). 2. The component derivatives are \(\frac{dx}{d\theta}=a\) and \(\frac{dy}{d\theta}=1+a\), so the tangent slope is \(\frac{1+a}{a}\). 3. Solving \(\frac{1+a}{a}=2\) gives \(a=1\), and this value keeps \(dx/d\theta\ne0\).

\(a=1\)

- What quantity does the radial rate describe? - Find how both Cartesian coordinates change with the angle. - Compare the proposed slope with the quotient of the Cartesian rates.

1. The radial rate \(\frac{dr}{d\theta}\) measures how radius changes with angle. It is not the Cartesian slope \(\frac{dy}{dx}\). 2. The Cartesian coordinate rates are \(\frac{dx}{d\theta}=-2(1+\cos\theta)\sin\theta\) and \(\frac{dy}{d\theta}=2\cos\theta+\cos(2\theta)\). 3. At \(\theta=\frac{\pi}{2}\), these are \(-2\) and \(-1\), so \(\frac{dy}{dx}=\frac{1}{2}\).

The radial rate is not Cartesian tangent slope. The actual slope is \(\frac{dy}{dx}=\frac{1}{2}\).

- What depends on the angle in \(x=r\cos\theta\)? - Check the same dependency when differentiating \(y=r\sin\theta\). - Use the corrected coordinate rates to find the slope.

1. Since the radius depends on \(\theta\), the product rule must include \(r'=3\cos\theta\). 2. The coordinate rates are \(\frac{dx}{d\theta}=r'\cos\theta-r\sin\theta=3\cos(2\theta)\) and \(\frac{dy}{d\theta}=r'\sin\theta+r\cos\theta=3\sin(2\theta)\). 3. At \(\theta=\frac{\pi}{6}\), the tangent slope is \(\frac{dy}{dx}=\frac{3\sin(\pi/3)}{3\cos(\pi/3)}=\sqrt{3}\).

Both \(r\) and the trigonometric factor depend on \(\theta\), so the product rule needs both terms. The correct slope is \(\sqrt{3}\).

- Differentiate the full rectangular coordinate products, including the radial factor. - Evaluate the horizontal and vertical rates separately at \(\theta=0\). - Determine the sign only after forming the quotient of those rates.

1. Since \(r'=e^{\theta}\), the rectangular derivatives are \(\frac{dx}{d\theta}=e^{\theta}(\cos(\theta)-\sin(\theta))\) and \(\frac{dy}{d\theta}=e^{\theta}(\sin(\theta)+\cos(\theta))\). 2. At \(\theta=0\), both derivatives equal \(1\). 3. Therefore, \(\frac{dy}{dx}=1\), not \(-1\).

The correct slope is \(\frac{dy}{dx}=1\).

- Derive one polar slope expression and use it at both angles. - Convert each polar location to rectangular coordinates independently. - Compare both the signs and magnitudes of the evaluated slopes.

1. The polar slope is \(\frac{dy}{dx}=\frac{\sin^{2}(\theta)-\cos(\theta)-\frac{1}{2}}{(1+\cos(\theta))\sin(\theta)}\). 2. At \(\theta=\frac{\pi}{4}\), the point is \(\left(\frac{1}{2}+\sqrt{2}, \frac{1}{2}+\sqrt{2}\right)\) and the slope is \(-2+\sqrt{2}\). 3. At \(\theta=\frac{\pi}{2}\), the point is \((0,2)\) and the slope is \(\frac{1}{2}\).

At \(\theta=\frac{\pi}{4}\): point \(\left(\frac{1}{2}+\sqrt{2}, \frac{1}{2}+\sqrt{2}\right)\), slope \(-2+\sqrt{2}\) At \(\theta=\frac{\pi}{2}\): point \((0,2)\), slope \(\frac{1}{2}\)

- Simplify one slope formula before substituting either angle. - Convert the polar points to Cartesian coordinates separately. - Compare the signs and sizes of the two slopes.

1. Since \(x=3\sin\theta\cos\theta\) and \(y=3\sin^{2}\theta\), the coordinate rates are \(\frac{dx}{d\theta}=3\cos(2\theta)\) and \(\frac{dy}{d\theta}=3\sin(2\theta)\). 2. Their ratio is \(\frac{dy}{dx}=\tan(2\theta)\) wherever the horizontal rate is nonzero. 3. At \(\theta=\frac{\pi}{6}\), the point is \(\left(\frac{3\sqrt{3}}{4},\frac{3}{4}\right)\), and the slope is \(\sqrt{3}\). 4. At \(\theta=\frac{\pi}{3}\), the point is \(\left(\frac{3\sqrt{3}}{4},\frac{9}{4}\right)\), and the slope is \(-\sqrt{3}\).

\(\frac{dy}{dx}=\tan(2\theta)\). At \(\theta=\frac{\pi}{6}\): \(\left(\frac{3\sqrt{3}}{4},\frac{3}{4}\right)\), slope \(\sqrt{3}\). At \(\theta=\frac{\pi}{3}\): \(\left(\frac{3\sqrt{3}}{4},\frac{9}{4}\right)\), slope \(-\sqrt{3}\).

- Use the same derivative formula at both symmetric angles. - Convert each polar point carefully, preserving the sign of \(\cos(\theta)\). - Keep the slope expression defined at the angles you need to evaluate.

1. The polar tangent slope simplifies to \(\frac{dy}{dx}=-\cot(2\theta)+\frac{1}{3\sin(2\theta)}\), which is defined at both requested angles. 2. At \(\theta=\frac{\pi}{4}\), the point is \(\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\) and the slope is \(\frac{1}{3}\). 3. At \(\theta=\frac{3\pi}{4}\), the point is \(\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\) and the slope is \(-\frac{1}{3}\).

At \(\theta=\frac{\pi}{4}\): point \(\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\), slope \(\frac{1}{3}\) At \(\theta=\frac{3\pi}{4}\): point \(\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)\), slope \(-\frac{1}{3}\)

- Derive one slope formula in terms of \(\theta\) and reuse it at both parameter values. - Remember that cosine is even and sine is odd when converting the negative-angle point. - Evaluate the radius separately at each angle before forming the rectangular coordinates.

1. The polar slope is \(\frac{dy}{dx}=\tan\left(\theta+\frac{\pi}{4}\right)\). 2. At \(\theta=-1\), the point is \(\left(\frac{\cos(1)}{e},-\frac{\sin(1)}{e}\right)\) and the slope is \(\tan\left(\frac{\pi}{4}-1\right)\). 3. At \(\theta=1\), the point is \((e\cos(1),e\sin(1))\) and the slope is \(\tan\left(\frac{\pi}{4}+1\right)\).

At \(\theta=-1\): point \(\left(\frac{\cos(1)}{e},-\frac{\sin(1)}{e}\right)\), slope \(\tan\left(\frac{\pi}{4}-1\right)\) At \(\theta=1\): point \((e\cos(1),e\sin(1))\), slope \(\tan\left(\frac{\pi}{4}+1\right)\)

- How is change in radius different from vertical Cartesian change? - Evaluate the horizontal and vertical Cartesian coordinate rates. - A horizontal tangent needs a zero numerator and a nonzero denominator.

1. The radial derivative is \(r'=-\frac{1}{\theta^{2}}\); this is not the Cartesian tangent slope. 2. The Cartesian slope formula is \(\frac{dy}{dx}=\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}\). 3. At \(\theta=1\), it becomes \(\frac{\cos(1)-\sin(1)}{-\cos(1)-\sin(1)}=\frac{\sin(1)-\cos(1)}{\sin(1)+\cos(1)}\). 4. This is not zero, so the tangent is not horizontal; the claim confuses radial change with Cartesian slope.

\(\frac{dy}{dx}=\frac{\sin(1)-\cos(1)}{\sin(1)+\cos(1)}\). The radial derivative alone cannot determine Cartesian tangent slope.

- A horizontal tangent is determined by the angular rate of the y-coordinate. - Substitute the angle before solving for the parameter. - Verify that the angular rate of the x-coordinate is nonzero.

1. A horizontal tangent requires \(\frac{dy}{d\theta}=r'\sin\theta+r\cos\theta=0\), with \(\frac{dx}{d\theta}\ne0\). 2. Here \(r'=-a\sin\theta\). 3. At \(\theta=\frac{\pi}{3}\), \(\frac{dy}{d\theta}=\frac{1-a}{2}\), so \(a=1\). 4. For \(a=1\), \(\frac{dx}{d\theta}=-\sqrt{3}\ne0\), confirming a horizontal tangent.

\(a=1\)

- Find the tangent direction from the coordinate rates. - The radial line points in the direction of the given polar angle. - Compare the two line directions using an angle relationship for slopes.

1. At \(\theta=\frac{\pi}{4}\), \(r=1\) and \(r'=2\). 2. The coordinate rates are \(\frac{dx}{d\theta}=\frac{\sqrt{2}}{2}\) and \(\frac{dy}{d\theta}=\frac{3\sqrt{2}}{2}\), so the tangent slope is \(3\). 3. The radial line has slope \(\tan\left(\frac{\pi}{4}\right)=1\). 4. If \(\phi\) is the acute angle between the lines, then \(\tan\phi=\left|\frac{3-1}{1+3}\right|=\frac{1}{2}\). 5. Therefore, \(\phi=\arctan\left(\frac{1}{2}\right)\).

\(\arctan\left(\frac{1}{2}\right)\)

- Differentiate both rectangular coordinate expressions with respect to the angle. - Preserve the signs of the coordinate rates to keep the correct orientation. - Normalize the resulting direction vector.

1. At \(\theta=\frac{\pi}{4}\), \(r=1\) and \(r'=-2\). 2. The coordinate rates are \(\frac{dx}{d\theta}=-\frac{3\sqrt{2}}{2}\) and \(\frac{dy}{d\theta}=-\frac{\sqrt{2}}{2}\). 3. This tangent vector is a positive scalar multiple of \(\langle-3,-1\rangle\). 4. Its magnitude is \(\sqrt{10}\), so the unit tangent vector is \(\frac{1}{\sqrt{10}}\langle-3,-1\rangle\).

\(\frac{1}{\sqrt{10}}\langle-3,-1\rangle\)

- Decompose the tangent direction into radial and angular components. - Perpendicularity to the radial direction removes one of those components. - Verify that the curve is not at the pole at the resulting angles.

1. The tangent vector has radial component \(r'\) and angular component \(r\). 2. It is perpendicular to the radial segment when the radial component is \(0\), provided \(r\ne0\). 3. Since \(r'=2\cos(2\theta)\), the solutions in \([0,\pi)\) are \(\theta=\frac{\pi}{4}\) and \(\theta=\frac{3\pi}{4}\). 4. The corresponding radii are \(3\) and \(1\), so both tangent vectors are nonzero.

\(\theta=\frac{\pi}{4}\) and \(\theta=\frac{3\pi}{4}\)

- Simplify the two coordinate rates before taking their quotient. - What tangent slope corresponds to a line parallel to \(y=x\)? - List every matching angle in the stated interval.

1. The Cartesian equations are \(x=\sin(2\theta)\) and \(y=2\sin^{2}\theta\). 2. Differentiate: \(\frac{dx}{d\theta}=2\cos(2\theta)\) and \(\frac{dy}{d\theta}=2\sin(2\theta)\), so \(\frac{dy}{dx}=\tan(2\theta)\) wherever defined. 3. A parallel tangent has slope \(1\); therefore, \(2\theta=\frac{\pi}{4}+k\pi\) for an integer \(k\). 4. On \([0,\pi)\), the solutions are \(\theta=\frac{\pi}{8}\) and \(\theta=\frac{5\pi}{8}\).

\(\frac{dy}{dx}=\tan(2\theta)\). Angles: \(\frac{\pi}{8}\) and \(\frac{5\pi}{8}\).

- Substitute the angle and radius into the polar slope formula. - Treat the unknown radial derivative as a variable. - Solve the resulting rational equation.

1. Let \(r'=q\). At \(\theta=\frac{\pi}{4}\), the polar slope becomes \(\frac{q+2}{q-2}\). 2. Set \(\frac{q+2}{q-2}=2\). 3. Solving \(q+2=2q-4\) gives \(q=6\).

\(\frac{dr}{d\theta}=6\)

- Use the angular rate of the x-coordinate for a vertical tangent. - Substitute the given angle before solving for the constant. - Verify that the y-coordinate rate is nonzero.

1. A vertical tangent requires \(\frac{dx}{d\theta}=r'\cos\theta-r\sin\theta=0\), with \(\frac{dy}{d\theta}\ne0\). 2. Since \(r'=1\), the condition at \(\theta=\frac{\pi}{4}\) is \(1-a-\frac{\pi}{4}=0\). 3. Thus \(a=1-\frac{\pi}{4}\). 4. For this value, \(\frac{dy}{d\theta}=\sqrt{2}\ne0\), confirming a vertical tangent.

\(a=1-\frac{\pi}{4}\)

- Compare the direction of the radius with the direction of motion along the curve. - Use the radial derivative at the specified angle. - A vector relationship can determine the angle without writing either line equation.

1. At the given angle, \(r=2\) and \(r'=6\cos(3\theta)=0\). 2. The radial direction vector is \(\langle\cos\theta,\sin\theta\rangle\). 3. When \(r'=0\), the tangent vector is proportional to \(\langle-\sin\theta,\cos\theta\rangle\). 4. These two direction vectors have dot product \(0\), so the angle between the lines is \(\frac{\pi}{2}\).

\(\frac{\pi}{2}\)

- Simplify the polar slope using the relationship between the radial function and its derivative. - Determine the slope perpendicular to the given line. - Solve the resulting trigonometric equation over the full interval.

1. Since \(r'=r\), the polar slope simplifies to \(\tan\left(\theta+\frac{\pi}{4}\right)\). 2. A line perpendicular to \(y=x\) has slope \(-1\). 3. Solving \(\tan\left(\theta+\frac{\pi}{4}\right)=-1\) gives \(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\) in the stated interval.

\(\theta=\frac{\pi}{2}\) and \(\theta=\frac{3\pi}{2}\)

- Evaluate the radius and radial derivative at the specified angle. - At the pole, use the rectangular coordinate rates directly. - Form the line through the origin with the resulting slope.

1. At \(\theta=1\), \(r=0\), so the point is \((0,0)\). 2. The radial derivative is \(r'=\frac{1}{\theta}\), so \(r'(1)=1\). 3. The coordinate rates are \(x'=\cos(1)\) and \(y'=\sin(1)\), giving slope \(\tan(1)\). 4. The tangent line through the origin is \(y=\tan(1)x\).

\(y=\tan(1)x\)

- With a negative radius, which Cartesian point does the stated angle reach? - Differentiate both Cartesian coordinates with respect to the angle. - Which coordinate rate must vanish for a horizontal tangent?

1. At \(\theta=\frac{3\pi}{2}\), \(r=-1\), so the Cartesian point is \((0,1)\). 2. The radial derivative is \(\frac{dr}{d\theta}=2\cos\theta=0\) at that angle. 3. Using \(x=r\cos\theta\) and \(y=r\sin\theta\), the coordinate rates are \(\frac{dx}{d\theta}=-1\) and \(\frac{dy}{d\theta}=0\). 4. The tangent is horizontal through \((0,1)\), so its equation is \(y=1\).

\(y=1\)

- Compare the two radial functions before calculating both slopes. - What happens to each Cartesian coordinate rate under a constant radial dilation? - Does the common factor change their ratio?

1. The second radial function is three times the first. 2. For the first curve, use \(r=1+\sin\theta\) and \(\frac{dr}{d\theta}=\cos\theta\) in the polar slope formula. The slope at \(\frac{\pi}{4}\) is \(-1-\sqrt{2}\). 3. For the second curve, both Cartesian coordinate rates are multiplied by \(3\), so their quotient is unchanged. 4. Both tangent slopes are \(-1-\sqrt{2}\).

Both slopes equal \(-1-\sqrt{2}\). Multiplying the radius by a positive constant does not change the tangent direction.

- Solve for the angles where the horizontal component derivative is zero. - Check the vertical component derivative at every candidate. - If both component derivatives vanish, examine the limiting reciprocal slope before deciding whether the tangent exists.

1. The component derivatives are \(\frac{dx}{d\theta}=2\cos(2\theta)-2\sin(\theta)\) and \(\frac{dy}{d\theta}=2\sin(2\theta)+2\cos(\theta)\). 2. Solving \(\frac{dx}{d\theta}=0\) gives \(\cos(2\theta)=\sin(\theta)\). Using \(\cos(2\theta)=1-2\sin^{2}(\theta)\) gives the candidates \(\theta=\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}\). 3. At \(\theta=\frac{\pi}{6}\) and \(\theta=\frac{5\pi}{6}\), \(\frac{dy}{d\theta}\ne0\), so both tangents are vertical. 4. At \(\theta=\frac{3\pi}{2}\), both component derivatives are zero. Applying l’Hôpital’s rule to \(\frac{dx/d\theta}{dy/d\theta}\) gives a limit of \(0\), so the cusp at the origin also has a vertical tangent.

\(\theta=\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}\)

- Set the vertical component derivative equal to zero, not the radial derivative. - Convert the resulting triple-angle equation into a polynomial in \(\cos(\theta)\). - Verify that the horizontal component derivative is nonzero at every candidate angle.

1. With \(r'=-6\sin(2\theta)\), the component derivatives simplify to \(\frac{dx}{d\theta}=18\sin^{3}(\theta)-15\sin(\theta)\) and \(\frac{dy}{d\theta}=\frac{3}{2}[3\cos(3\theta)-\cos(\theta)]\). 2. Setting \(\frac{dy}{d\theta}=0\) and using \(\cos(3\theta)=4\cos^{3}(\theta)-3\cos(\theta)\) gives \(2\cos(\theta)(6\cos^{2}(\theta)-5)=0\). 3. On \([0,\pi)\), the candidates are \(\theta=\arccos\left(\sqrt{\frac{5}{6}}\right),\frac{\pi}{2},\pi-\arccos\left(\sqrt{\frac{5}{6}}\right)\). 4. At each candidate, \(\frac{dx}{d\theta}\ne0\), so all three give horizontal tangents.

\(\theta=\arccos\left(\sqrt{\frac{5}{6}}\right),\frac{\pi}{2},\pi-\arccos\left(\sqrt{\frac{5}{6}}\right)\)

- Set the horizontal component derivative equal to zero for vertical tangency. - Use the triple-angle identity to obtain an algebraic equation in \(\cos(\theta)\). - Check that the vertical component derivative does not vanish at the candidate angles.

1. With \(r'=6\cos(2\theta)\), the component derivatives simplify to \(\frac{dx}{d\theta}=\frac{3}{2}[\cos(\theta)+3\cos(3\theta)]\) and \(\frac{dy}{d\theta}=-18\sin^{3}(\theta)+12\sin(\theta)\). 2. Setting \(\frac{dx}{d\theta}=0\) and using \(\cos(3\theta)=4\cos^{3}(\theta)-3\cos(\theta)\) gives \(6\cos(\theta)(3\cos^{2}(\theta)-2)=0\). 3. On \([0,\pi)\), the candidates are \(\theta=\arccos\left(\sqrt{\frac{2}{3}}\right),\frac{\pi}{2},\pi-\arccos\left(\sqrt{\frac{2}{3}}\right)\). 4. At each candidate, \(\frac{dy}{d\theta}\ne0\), so all three give vertical tangents.

\(\theta=\arccos\left(\sqrt{\frac{2}{3}}\right),\frac{\pi}{2},\pi-\arccos\left(\sqrt{\frac{2}{3}}\right)\)

- Solve for the angles where the vertical component derivative is zero. - Check the horizontal component derivative at every candidate. - If both component derivatives vanish, examine the limiting slope before deciding whether the tangent exists.

1. The component derivatives are \(\frac{dx}{d\theta}=-\sin(\theta)-\sin(2\theta)\) and \(\frac{dy}{d\theta}=\cos(\theta)+\cos(2\theta)\). 2. Solving \(\frac{dy}{d\theta}=0\) gives \(\cos(\theta)+\cos(2\theta)=0\), so the candidates are \(\theta=\frac{\pi}{3},\pi,\frac{5\pi}{3}\). 3. At \(\theta=\frac{\pi}{3}\) and \(\theta=\frac{5\pi}{3}\), \(\frac{dx}{d\theta}\ne0\), so both tangents are horizontal. 4. At \(\theta=\pi\), both component derivatives are zero. Applying l’Hôpital’s rule to \(\frac{dy/d\theta}{dx/d\theta}\) gives a limiting slope of \(0\), so the cusp at the origin also has a horizontal tangent.

\(\theta=\frac{\pi}{3},\pi,\frac{5\pi}{3}\)

- Express the slope as the quotient of the two Cartesian rates. - A derivative with respect to \(\theta\) is not yet a derivative with respect to \(x\). - Check the sign of the Cartesian second derivative.

1. The Cartesian rates are \(\frac{dx}{d\theta}=-2\sin(2\theta)\) and \(\frac{dy}{d\theta}=2\cos(2\theta)\), giving \(\frac{dy}{dx}=-\cot(2\theta)\). 2. Differentiate the slope: \(\frac{d}{d\theta}\left(\frac{dy}{dx}\right)=2\csc^{2}(2\theta)\). 3. Divide by the horizontal rate: \(\frac{d^{2}y}{dx^{2}}=-\frac{1}{\sin^{3}(2\theta)}\). 4. At \(\theta=\frac{\pi}{4}\), this is \(-1\), so the curve is concave down.

\(\frac{dy}{dx}=-\cot(2\theta)\). \(\frac{d}{d\theta}\left(\frac{dy}{dx}\right)=2\csc^{2}(2\theta)\). At \(\theta=\frac{\pi}{4}\), \(\frac{d^{2}y}{dx^{2}}=-1\); concave down.

- Keep \(\theta\) as the parameter when computing both derivative stages. - Differentiate the slope function before dividing by the horizontal component derivative. - Evaluate the final expression at the angle and use its sign for concavity.

1. The component derivatives are \(\frac{dx}{d\theta}=-\sin(\theta)+\cos(2\theta)\) and \(\frac{dy}{d\theta}=\sin(2\theta)+\cos(\theta)\). 2. Thus, \(\frac{dy}{dx}=\frac{(2\sin(\theta)+1)\cos(\theta)}{-\sin(\theta)+\cos(2\theta)}\). 3. Differentiating this slope with respect to \(\theta\) and dividing by \(dx/d\theta\) gives \(\frac{d^{2}y}{dx^{2}}=-\frac{3}{(1+\sin(\theta))^{2}(2\sin(\theta)-1)^{3}}\). 4. At \(\theta=0\), the value is \(3\), so the curve is concave up.

\(\frac{d^{2}y}{dx^{2}}=3\); concave up

- Compute the first polar derivative from the two rectangular component derivatives. - Differentiate the quotient with respect to \(\theta\), then divide by \(dx/d\theta\). - Substitute the angle only after obtaining the second-derivative expression.

1. The component derivatives are \(\frac{dx}{d\theta}=e^{\theta}(\cos(\theta)-\sin(\theta))\) and \(\frac{dy}{d\theta}=e^{\theta}(\sin(\theta)+\cos(\theta))\). 2. Hence, \(\frac{dy}{dx}=\frac{\sin(\theta)+\cos(\theta)}{\cos(\theta)-\sin(\theta)}\). 3. Differentiating the slope and dividing by \(dx/d\theta\) gives \(\frac{d^{2}y}{dx^{2}}=\frac{2e^{-\theta}}{(\cos(\theta)-\sin(\theta))^{3}}\). 4. At \(\theta=0\), the value is \(2\), so the curve is concave up.

\(\frac{d^{2}y}{dx^{2}}=2\); concave up

- Denote the rectangular component derivatives by two temporary functions to organize the quotient calculation. - Differentiate the first-derivative quotient with respect to \(\theta\), then divide by the horizontal rate. - Evaluate all needed component derivatives at the specified angle before simplifying.

1. Let \(A=\frac{dx}{d\theta}\) and \(B=\frac{dy}{d\theta}\). At \(\theta=\frac{\pi}{6}\), \(A=\frac{3\sqrt{3}}{4}\) and \(B=\frac{15}{4}\). 2. Differentiate the component rates: \(A'=\frac{d^{2}x}{d\theta^{2}}=-\frac{57}{4}\) and \(B'=\frac{d^{2}y}{d\theta^{2}}=-\frac{3\sqrt{3}}{4}\). 3. Since \(\frac{dy}{dx}=\frac{B}{A}\), \(\frac{d^{2}y}{dx^{2}}=\frac{1}{A}\frac{d}{d\theta}\left(\frac{B}{A}\right)=\frac{B'A-BA'}{A^{3}}\). 4. Substitution gives \(\frac{d^{2}y}{dx^{2}}=\frac{368\sqrt{3}}{27}>0\), so the curve is concave up.

\(\frac{d^{2}y}{dx^{2}}=\frac{368\sqrt{3}}{27}\); concave up

- Find the rectangular point and tangent slope at the specified angle. - Relate the tangent point to the y-intercept form of a line. - Simplify only after substituting the exact point and slope.

1. The point on the curve is \(\left(\frac{\pi\sqrt{2}}{8},\frac{\pi\sqrt{2}}{8}\right)\). 2. With \(r'=1\), the tangent slope at \(\theta=\frac{\pi}{4}\) is \(m=\frac{4+\pi}{4-\pi}\). 3. In \(y=mx+b\), substitute the point to get \(b=\frac{\pi\sqrt{2}}{8}(1-m)\). 4. Simplifying gives \(b=-\frac{\pi^2\sqrt{2}}{4(4-\pi)}\).

\(b=-\frac{\pi^2\sqrt{2}}{4(4-\pi)}\)

- Determine the polar point and the normal direction first. - Rewrite the normal line in standard form. - Use the perpendicular distance from a point to a line.

1. The point on the curve is \((0,\pi)\). 2. The tangent slope is \(-\frac{2}{\pi}\), so the normal slope is \(\frac{\pi}{2}\). 3. The normal line is \(y-\pi=\frac{\pi}{2}x\), or \(\pi x-2y+2\pi=0\). 4. The distance from \((0,0)\) to this line is \(\frac{2\pi}{\sqrt{\pi^2+4}}\).

\(\frac{2\pi}{\sqrt{\pi^2+4}}\)

- Find the horizontal-tangent angles before comparing their points. - Convert each polar point to an x-coordinate. - The signs of the cosine values distinguish the two candidates quickly.

1. A horizontal tangent requires \(r'\sin\theta+r\cos\theta=r(\cos\theta-\sin\theta)=0\). 2. The two angles are \(\theta=\frac{\pi}{4}\) and \(\theta=\frac{5\pi}{4}\). 3. Their x-coordinates are \(\frac{e^{-\pi/4}}{\sqrt{2}}\) and \(-\frac{e^{-5\pi/4}}{\sqrt{2}}\), respectively. 4. The larger x-coordinate occurs at \(\theta=\frac{\pi}{4}\), where the point is \(\left(\frac{e^{-\pi/4}}{\sqrt{2}},\frac{e^{-\pi/4}}{\sqrt{2}}\right)\).

\(\theta=\frac{\pi}{4}\), at \(\left(\frac{e^{-\pi/4}}{\sqrt{2}},\frac{e^{-\pi/4}}{\sqrt{2}}\right)\)

- Express the tangent direction using radial and perpendicular components. - Compare the radial function with its derivative. - A constant component ratio produces a constant angle.

1. In radial and angular directions, the tangent vector has components \(r'\) and \(r\). 2. Since \(r'=2e^{2\theta}=2r\), the tangent vector has proportional components \(2r\) radially and \(r\) angularly. 3. If \(\phi\) is the acute angle from the radial direction to the tangent, then \(\tan\phi=\frac{r}{r'}=\frac{1}{2}\). 4. Thus \(\phi=\arctan\left(\frac{1}{2}\right)\), independent of \(\theta\).

\(\phi=\arctan\left(\frac{1}{2}\right)\), and it is constant.

- Find both the tangent slope and its rate of change with respect to x at the base point. - View the tangent slope itself as a function to be approximated. - Use the small change in x from the base point.

1. At \(\theta=0\), the point is \((2,0)\) and the tangent slope is \(\frac{dy}{dx}=2\). 2. Differentiating the polar slope with respect to \(\theta\) and dividing by \(\frac{dx}{d\theta}\) gives \(\frac{d^2y}{dx^2}=6\) at the base point. 3. Treat the tangent slope as a function of x and linearize near \(x=2\): \(m(x)\approx2+6(x-2)\). 4. At \(x=2.05\), \(m(2.05)\approx2+6(0.05)=2.3\).

The estimated tangent slope is \(2.3\).

- Use the rectangular point condition to determine the radius at the angle. - Evaluate the polar slope formula at the same angle. - Solve the point and slope conditions in sequence.

1. At \(\theta=0\), the radius is \(a\), so the point condition gives \(a=2\). 2. The radial derivative is \(r'=b\cos\theta\), so \(r'(0)=b\). 3. At \(\theta=0\), the polar slope is \(\frac{r}{r'}=\frac{a}{b}\). 4. Solving \(\frac{2}{b}=\frac{1}{2}\) gives \(b=4\).

\(a=2\) and \(b=4\)

- First express the rectangular coordinate rates using the radial data. - Differentiate those rates once more with respect to the angle. - Convert the angular change in slope to a derivative with respect to x.

1. At \(\theta=0\), \(x'=r'=1\) and \(y'=r=2\). 2. The second angular derivatives are \(x''=r''-r=-2\) and \(y''=2r'=2\). 3. The angular derivative of the slope is \(\frac{y''x'-y'x''}{(x')^2}=\frac{2(1)-2(-2)}{1}=6\). 4. Dividing by \(x'=1\) gives \(\frac{d^2y}{dx^2}=6\).

\(\frac{d^2y}{dx^2}=6\)

- Express the first and second rectangular coordinate rates from the radial data. - Use the parametric second-derivative relationship. - Track the sign of the cubed horizontal rate carefully.

1. At \(\theta=\frac{\pi}{2}\), the coordinate rates are \(x'=-r=-1\) and \(y'=r'=2\). 2. The second angular derivatives are \(x''=-2r'=-4\) and \(y''=r''-r=-2\). 3. The second derivative is \(\frac{y''x'-y'x''}{(x')^3}\). 4. Substitution gives \(\frac{(-2)(-1)-2(-4)}{(-1)^3}=-10\).

\(\frac{d^2y}{dx^2}=-10\)

- Write the tangent slope before substituting the pole value. - Simplify the expression for nonzero angles near the pole. - Use a limit to determine the tangent direction.

1. With \(r=\theta^2\) and \(r'=2\theta\), the polar slope is \(\frac{dy}{dx}=\frac{2\theta\sin\theta+\theta^2\cos\theta}{2\theta\cos\theta-\theta^2\sin\theta}\). 2. For \(\theta>0\), cancel a factor of \(\theta\) to obtain \(\frac{2\sin\theta+\theta\cos\theta}{2\cos\theta-\theta\sin\theta}\). 3. Taking \(\theta\to0^+\) gives limiting slope \(0\). 4. The curve reaches \((0,0)\), so the tangent line at the pole is \(y=0\).

The limiting slope is \(0\), and the tangent line is \(y=0\).

- Express the second derivative at the specified angle in terms of the parameter. - The positivity condition will select one root. - Verify that the horizontal coordinate rate is nonzero.

1. At \(\theta=0\), the rectangular coordinate rates give \(x'=1\) and \(y'=a\). 2. Differentiating the polar slope and dividing by \(x'\) gives \(\frac{d^2y}{dx^2}=a^2+2\) at \(\theta=0\). 3. Thus \(a^2+2=11\), so \(a^2=9\). 4. Because \(a>0\), \(a=3\).

\(a=3\)

- Differentiate the implicit radial equation twice. - Evaluate the radial derivatives at the specified angle. - Use the rectangular parametric second-derivative relationship.

1. At \(\theta=\frac{\pi}{2}\), \(r=2\). 2. Differentiating \(r^3=8\sin\theta\) gives \(3r^2r'=8\cos\theta\), so \(r'=0\). 3. Differentiating again gives \(6r(r')^2+3r^2r''=-8\sin\theta\), so \(r''=-\frac{2}{3}\). 4. The parametric second derivative evaluates to \(-\frac{2}{3}\), so the curve is concave down.

\(\frac{d^2y}{dx^2}=-\frac{2}{3}\); the curve is concave down.

- Express both the position vector and tangent vector using radial and angular directions. - A line through the point and the origin would require those two vectors to be parallel. - Test parallelism with a planar determinant.

1. The position vector is \(\mathbf{p}=r\langle\cos\theta,\sin\theta\rangle\). 2. A tangent direction is \(\mathbf{p}'=r'\langle\cos\theta,\sin\theta\rangle+r\langle-\sin\theta,\cos\theta\rangle\). 3. The determinant of these vectors is \(\det(\mathbf{p},\mathbf{p}')=r^2\). 4. Since \(r\ne0\), this determinant is nonzero, so the tangent direction is not parallel to the position vector. Therefore, the tangent line cannot contain the origin.

At a regular point with \(r\ne0\), \(\det(\mathbf{p},\mathbf{p}')=r^2\ne0\). Thus the tangent direction is not parallel to the radial position vector, so the tangent line cannot pass through the origin.

- Turn the polar equation into separate \(x\)- and \(y\)-coordinates depending on \(\theta\). - What does the first derivative say about the tangent, and what does it not say about the curve's shape? - When finding the second Cartesian derivative, account for how \(x\) changes with \(\theta\).

1. Write \(x=\theta\cos\theta\) and \(y=\theta\sin\theta\). Then \(\frac{dx}{d\theta}=\cos\theta-\theta\sin\theta\) and \(\frac{dy}{d\theta}=\sin\theta+\theta\cos\theta\). 2. At \(\theta=0\), these rates are \(1\) and \(0\), so \(\frac{dy}{dx}=0\) and the tangent is horizontal. 3. The second component derivatives are \(\frac{d^2x}{d\theta^2}=-2\sin\theta-\theta\cos\theta\) and \(\frac{d^2y}{d\theta^2}=2\cos\theta-\theta\sin\theta\). At zero they are \(0\) and \(2\). 4. Differentiating \(\frac{dy}{dx}\) with respect to \(\theta\), then dividing by \(\frac{dx}{d\theta}\), gives the right-hand value \(\frac{d^2y}{dx^2}=2\) at the pole. The curve bends upward just to the right of the pole, so Leonie is incorrect.

From the right, \(\frac{dy}{dx}=0\) and \(\frac{d^2y}{dx^2}=2\) at \(\theta=0\). Leonie is incorrect: a horizontal tangent does not mean the curve is locally straight.

- Use the stated positive-radius branch when solving for \(r\). - Differentiate the polar relation implicitly to find the radial rate. - Check the two Cartesian coordinate rates for tangent direction.

1. At \(\theta=\frac{\pi}{6}\), \(r=\sqrt{2}\), giving the point \(\left(\frac{\sqrt{6}}{2},\frac{\sqrt{2}}{2}\right)\). 2. Implicit differentiation gives \(2r\frac{dr}{d\theta}=-8\sin(2\theta)\), so \(\frac{dr}{d\theta}=-\sqrt{6}\) at this angle. 3. The Cartesian coordinate rates are \(\frac{dx}{d\theta}=-2\sqrt{2}\) and \(\frac{dy}{d\theta}=0\). 4. The tangent is horizontal through the point, so \(y=\frac{\sqrt{2}}{2}\).

\(y=\frac{\sqrt{2}}{2}\)

- Find the derivative of the radius, then differentiate both Cartesian coordinates. - The same Cartesian point can be reached with different tangent directions. - Check which component rate is zero at each angle.

1. Since \(r=\sin(2\theta)\), \(\frac{dr}{d\theta}=2\cos(2\theta)\). Also, \(\frac{dx}{d\theta}=r'\cos\theta-r\sin\theta\) and \(\frac{dy}{d\theta}=r'\sin\theta+r\cos\theta\), where \(r'=\frac{dr}{d\theta}\). 2. At \(\theta=0\), \(r=0\) and \(r'=2\), giving rates \(2\) and \(0\); the tangent line is \(y=0\). 3. At \(\theta=\frac{\pi}{2}\), \(r=0\) and \(r'=-2\), giving rates \(0\) and \(-2\); the tangent line is \(x=0\). 4. These two visits to the same point have different tangent directions.

At \(\theta=0\): \(\frac{dx}{d\theta}=2\), \(\frac{dy}{d\theta}=0\), tangent \(y=0\). At \(\theta=\frac{\pi}{2}\): \(\frac{dx}{d\theta}=0\), \(\frac{dy}{d\theta}=-2\), tangent \(x=0\).

- A vertical tangent occurs where the horizontal coordinate rate vanishes but the vertical rate does not. - Solve the trig equation on the full stated interval. - Convert the qualifying angles to points before finding the line through them.

1. Differentiate \(x=(2-\cos\theta)\cos\theta\) to obtain \(\frac{dx}{d\theta}=2\sin\theta(\cos\theta-1)\). 2. This rate is zero at \(\theta=0\) and \(\theta=\pi\); the vertical-coordinate rate is nonzero at both, so these are vertical tangents. 3. The points are \((1,0)\) and \((-3,0)\). 4. The line through both is \(y=0\).

\(\frac{dx}{d\theta}=2\sin\theta(\cos\theta-1)\). Vertical-tangent points: \((1,0)\), \((-3,0)\). Line: \(y=0\).