Free math worksheets

- Convert the polar relation into parametric coordinates using \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\). - Differentiate both products with respect to \(\theta\), including \(r'(\theta)\). - Evaluate the two component derivatives before taking their quotient.

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

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

- View \(\theta\) as the parameter in the rectangular coordinate functions. - Apply the product rule to both \(r\cos(\theta)\) and \(r\sin(\theta)\). - Simplify with double-angle identities before substituting the angle.

1. Use \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\), with \(r'=3\cos(\theta)\). 2. The derivatives simplify to \(\frac{dx}{d\theta}=3\cos(2\theta)\) and \(\frac{dy}{d\theta}=3\sin(2\theta)\). 3. At \(\theta=\frac{\pi}{6}\), the slope is \(\frac{dy}{dx}=\frac{3\sin(\pi/3)}{3\cos(\pi/3)}=\sqrt{3}\).

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

- Write the polar curve as a parametric curve in \(x\) and \(y\). - Differentiate the products before substituting the requested angle. - Form \(\frac{dy}{dx}\) as \(\frac{dy/d\theta}{dx/d\theta}\), checking that the denominator is nonzero.

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

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

- Differentiate \(r(\theta)\) with the chain rule before using the polar slope formulas. - Evaluate \(r\) and \(r'\) at the given angle to reduce algebra. - Compute the rectangular component derivatives separately, then divide.

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, \(\frac{dy}{dx}=\frac{5/2}{\sqrt{3}/2}=\frac{5\sqrt{3}}{3}\).

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

- Treat the polar equation as a parametric curve in \(x\) and \(y\). - Differentiate the products using both \(r\) and \(r'\). - Evaluate the component derivatives at \(\theta=0\) before forming their quotient.

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 using the polar slope formulas. - Simplify the rectangular component derivatives algebraically before substituting the angle. - Divide \(dy/d\theta\) by \(dx/d\theta\), retaining the sign of each value.

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

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

- Substitute the tabulated radial value and radial rate into both rectangular derivative formulas. - Use the exact sine and cosine values at the listed angle. - Check the horizontal rate before dividing to distinguish a finite slope from a vertical tangent.

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 the radial value and radial derivative in the rectangular rate formulas. - Substitute the trigonometric values at \(\frac{\pi}{2}\) before simplifying signs. - Divide only after confirming the horizontal component derivative is nonzero.

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, \(\frac{dy}{dx}=\frac{-2}{-1}=2\).

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

- Substitute the listed radial data into both component derivative formulas. - Evaluate the sine and cosine factors at \(\theta=\pi\). - A zero horizontal rate with a nonzero vertical rate indicates a vertical tangent.

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.

- Look for a standard rectangular coordinate inside the polar equation. - Rewrite the equation before applying a derivative formula. - Use the geometry of the resulting rectangular graph.

1. Multiply by \(\sin\theta\) to obtain \(r\sin\theta=2\). 2. Since \(y=r\sin\theta\), the rectangular equation is \(y=2\). 3. This is a horizontal line, so its tangent slope is \(0\) at every point of the curve.

The graph is the horizontal line \(y=2\), and its tangent slope is \(0\).

- 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 \(\langle0,1\rangle\).

\(\langle0,1\rangle\)

- A radial derivative is not the same as a rectangular tangent slope. - Evaluate both rectangular coordinate rates. - Classify the tangent from which rate is zero.

1. At \(\theta=0\), \(r=1\) and \(r'=0\), so the point is \((1,0)\). 2. The rectangular 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, not horizontal. 4. Its equation is \(x=1\), so the student’s claim is false.

The claim is false. At \(\theta=0\), the curve has the vertical tangent \(x=1\).

- Convert the specified polar point to rectangular coordinates first. - Use the polar derivative formulas to determine the tangent slope. - Write the line through the calculated point using the resulting slope.

1. At \(\theta=\frac{\pi}{4}\), \(r=\sqrt{2}\), so the rectangular point is \((x,y)=(1,1)\). 2. With \(r'=-2\sin(\theta)\), the polar component derivatives at this angle are \(\frac{dx}{d\theta}=-2\) and \(\frac{dy}{d\theta}=0\). 3. Thus, the tangent slope is \(0\), and the tangent line through \((1,1)\) is \(y=1\).

\(y=1\)

- Convert the polar point to rectangular coordinates. - Find the tangent slope from the polar component derivatives. - Use the perpendicular slope to write the normal line through the point.

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

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

- Evaluate the radial function and convert the point to rectangular coordinates. - Compute \(r'\), then use it in both polar component derivatives. - Use the slope and point in point-slope form without prematurely expanding the line.

1. At \(\theta=\frac{\pi}{3}\), \(r=\frac{3\sqrt{3}}{2}\), so the rectangular point is \(\left(\frac{3\sqrt{3}}{4}, \frac{9}{4}\right)\). 2. With \(r'=3\cos(\theta)\), the component derivatives at this angle 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. Therefore, the tangent line is \(y-\frac{9}{4}=-\sqrt{3}\left(x-\frac{3\sqrt{3}}{4}\right)\).

\(y-\frac{9}{4}=-\sqrt{3}\left(x-\frac{3\sqrt{3}}{4}\right)\)

- Convert the specified polar point to rectangular coordinates. - Compute the tangent slope from the two polar component derivatives. - Use the negative reciprocal of the tangent slope for the normal line.

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 \(dx/d\theta\ne0\).

\(a=2\)

- Find the values of the radial function and its derivative at \(\theta=0\). - Substitute those values into the polar formulas for \(dx/d\theta\) and \(dy/d\theta\). - Solve the slope equation and reject any value that makes the horizontal rate zero.

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 \(r\) and \(r'\) at \(\theta=\frac{\pi}{2}\) before doing algebra. - Compute the two rectangular rates and simplify their quotient. - Match that quotient to the prescribed slope and check the denominator condition.

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.

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

- Evaluate the exponential radial function and its derivative at the given angle. - Use both values in the rectangular component derivative formulas. - Solve the resulting rational equation and confirm the slope is defined.

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 \(dx/d\theta\ne0\).

\(a=1\)

- Distinguish a change in radial distance from a change in rectangular coordinates. - Differentiate the full expressions \(x=r\cos(\theta)\) and \(y=r\sin(\theta)\). - Evaluate both component derivatives before taking their quotient.

1. The radial rate \(dr/d\theta\) measures how the distance from the origin changes; it is not the rectangular slope \(dy/dx\). 2. The rectangular derivatives 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 the rectangular tangent slope. The actual slope is \(\frac{dy}{dx}=\frac{1}{2}\).

- Identify every factor in \(x=r\cos(\theta)\) that depends on \(\theta\). - Apply the product rule to both rectangular coordinate functions. - Evaluate the two component derivatives before taking their quotient.

1. The radial function depends on \(\theta\), so the product rule must include \(r'=3\cos(\theta)\). 2. The rectangular derivatives 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 slope is \(\frac{dy}{dx}=\frac{3\sin(\pi/3)}{3\cos(\pi/3)}=\sqrt{3}\).

The student must differentiate both \(r\) and the trigonometric factor. The actual slope is \(\frac{dy}{dx}=\sqrt{3}\).

- 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 the polar slope once before evaluating either angle. - Compute the radius separately at each angle before converting to \((x,y)\). - Notice whether the two slopes have equal magnitude or equal sign.

1. The polar slope simplifies to \(\frac{dy}{dx}=\tan(2\theta)\). 2. At \(\theta=\frac{\pi}{6}\), the point is \(\left(\frac{3\sqrt{3}}{4}, \frac{3}{4}\right)\) and the slope is \(\sqrt{3}\). 3. At \(\theta=\frac{\pi}{3}\), the point is \(\left(\frac{3\sqrt{3}}{4}, \frac{9}{4}\right)\) and the slope is \(-\sqrt{3}\).

At \(\theta=\frac{\pi}{6}\): point \(\left(\frac{3\sqrt{3}}{4}, \frac{3}{4}\right)\), slope \(\sqrt{3}\) At \(\theta=\frac{\pi}{3}\): point \(\left(\frac{3\sqrt{3}}{4}, \frac{9}{4}\right)\), slope \(-\sqrt{3}\)

- Use the same derivative formula at both symmetric angles. - Convert each polar point carefully, preserving the sign of \(\cos(\theta)\). - Compare how the symmetry of the points is reflected in the slopes.

1. The polar slope simplifies to \(\frac{dy}{dx}=-\frac{1}{\tan(2\theta)}+\frac{1}{3\sin(2\theta)}\). 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 \(\tan\left(\frac{\pi}{4}+1\right)\)

- Distinguish change in radius from change in the rectangular y-coordinate. - Evaluate both rectangular coordinate rates at the angle. - A horizontal tangent requires a zero numerator and a nonzero denominator.

1. A radial derivative is not the rectangular tangent slope. Here \(r'=-\frac{1}{\theta^2}\). 2. The rectangular slope is \(\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}\). 3. At \(\theta=1\), this becomes \(\frac{\cos(1)-\sin(1)}{-\cos(1)-\sin(1)}=\frac{\sin(1)-\cos(1)}{\sin(1)+\cos(1)}\). 4. The slope is not \(0\), so the tangent is not horizontal; the student confused radial change with rectangular slope.

The actual slope is \(\frac{\sin(1)-\cos(1)}{\sin(1)+\cos(1)}\). The claim is false because \(r'\) alone is not \(\frac{dy}{dx}\).

- 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 polar derivative for this particular curve. - Match the tangent slope to the slope of the given line. - Include every angle in the specified interval.

1. The rectangular parameterization gives tangent slope \(\frac{dy}{dx}=\tan(2\theta)\). 2. Parallelism to \(y=x\) requires \(\tan(2\theta)=1\). 3. Thus \(2\theta=\frac{\pi}{4}+k\pi\). 4. The solutions in \([0,\pi)\) are \(\theta=\frac{\pi}{8}\) and \(\theta=\frac{5\pi}{8}\).

\(\theta=\frac{\pi}{8}\) and \(\theta=\frac{5\pi}{8}\)

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

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

- Convert the polar coordinates using the signed radius. - Evaluate both rectangular coordinate rates at the angle. - Use the zero and nonzero rates to classify the tangent.

1. At \(\theta=\frac{3\pi}{2}\), \(r=-1\), so the rectangular point is \((0,1)\). 2. The radial derivative is \(r'=2\cos\theta=0\). 3. The angular coordinate rates are \(\frac{dx}{d\theta}=-1\) and \(\frac{dy}{d\theta}=0\), so the tangent is horizontal. 4. The tangent line through \((0,1)\) is \(y=1\).

\(y=1\)

- Notice that one radial function is a constant multiple of the other. - Compare how that constant affects the two rectangular coordinate rates. - A common nonzero factor cancels in their ratio.

1. The second curve is a dilation of the first by a factor of \(3\). 2. For the first curve, substituting \(r=1+\sin\theta\) and \(r'=\cos\theta\) into the polar slope formula gives \(-1-\sqrt{2}\) at \(\theta=\frac{\pi}{4}\). 3. For the second curve, both rectangular coordinate rates are multiplied by \(3\), so their ratio is unchanged. 4. Both tangent slopes are \(-1-\sqrt{2}\).

Both slopes are \(-1-\sqrt{2}\). A radial dilation scales both coordinate rates equally, so it does not change tangent direction.

- 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 \(\frac{dx}{d\theta}=-2\sin(\theta)-2\sin(2\theta)\) and \(\frac{dy}{d\theta}=2\cos(\theta)+2\cos(2\theta)\). 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.

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

- 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.

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

- First compute the polar slope as a function of \(\theta\). - Differentiate that slope with respect to \(\theta\), then divide by \(dx/d\theta\). - Use the sign of the evaluated second derivative to determine concavity.

1. The rectangular component derivatives are \(\frac{dx}{d\theta}=-2\sin(2\theta)\) and \(\frac{dy}{d\theta}=2\cos(2\theta)\), so \(\frac{dy}{dx}=-\cot(2\theta)\). 2. Differentiate the first derivative with respect to \(\theta\): \(\frac{d}{d\theta}\left(\frac{dy}{dx}\right)=2\csc^{2}(2\theta)\). 3. Therefore, \(\frac{d^{2}y}{dx^{2}}=\frac{2\csc^{2}(2\theta)}{-2\sin(2\theta)}=-\frac{1}{\sin^{3}(2\theta)}\). 4. At \(\theta=\frac{\pi}{4}\), the value is \(-1\), so the curve is concave down.

\(\frac{d^{2}y}{dx^{2}}=-1\); 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.

- A zero first derivative determines only the tangent direction. - Use the rate of change of the tangent slope to test whether the curve bends. - Relate the first two derivatives to a local quadratic approximation.

1. The rectangular parameterization is \(x=\theta\cos\theta\) and \(y=\theta\sin\theta\). 2. At \(\theta=0\), the point is \((0,0)\) and \(\frac{dy}{dx}=0\). 3. Differentiating the slope with respect to \(\theta\) and dividing by \(\frac{dx}{d\theta}\) gives \(\frac{d^2y}{dx^2}=2\) at the pole. 4. Therefore, the curve bends upward and has local quadratic model \(y\approx x^2\), not a straight-line shape.

The claim is false. At the pole, \(\frac{d^2y}{dx^2}=2\), and the local quadratic model is \(y\approx x^2\).

- Determine the signed radial value on the stated branch. - Differentiate the polar relation implicitly to find the radial rate. - Use the two rectangular coordinate rates to classify the tangent.

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

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

- Evaluate the rectangular coordinate rates at each visit to the pole. - A pole can have different tangent directions for different parameter values. - Classify each tangent from which coordinate rate vanishes.

1. At \(\theta=0\), \(r=0\) and \(r'=2\). The coordinate rates are \(x'=2\) and \(y'=0\), so the tangent is \(y=0\). 2. At \(\theta=\frac{\pi}{2}\), \(r=0\) and \(r'=-2\). The coordinate rates are \(x'=0\) and \(y'=-2\), so the tangent is \(x=0\). 3. The same point therefore has two distinct tangent lines from the two branches.

At \(\theta=0\), the tangent is \(y=0\). At \(\theta=\frac{\pi}{2}\), the tangent is \(x=0\).

- Locate the vertical tangents using the horizontal coordinate rate. - Convert each resulting polar point to rectangular coordinates. - Use the two points to identify the joining line.

1. A vertical tangent requires \(\frac{dx}{d\theta}=r'\cos\theta-r\sin\theta=2\sin\theta(\cos\theta-1)=0\). 2. The valid angles are \(\theta=0\) and \(\theta=\pi\); the corresponding y-coordinate rates are nonzero. 3. The points are \((1,0)\) and \((-3,0)\). 4. The chord through these points is the x-axis, with equation \(y=0\).

\(y=0\)

- The horizontal tangent makes the first-order change in y equal to zero. - Find how the tangent slope changes with respect to x at the given angle. - Use the local quadratic change associated with the specified horizontal displacement.

1. At \(\theta=\frac{\pi}{4}\), \(\frac{dy}{dx}=0\). 2. Differentiating the polar slope with respect to \(\theta\) and dividing by \(\frac{dx}{d\theta}\) gives \(\frac{d^2y}{dx^2}=-\frac{3\sqrt{2}}{8}\). 3. With \(\Delta x=\frac{1}{5}\), the quadratic estimate is \(y\approx\sqrt{2}+\frac{1}{2}\left(-\frac{3\sqrt{2}}{8}\right)\left(\frac{1}{5}\right)^2\). 4. Therefore, \(y\approx\frac{397\sqrt{2}}{400}\).

\(y\approx\frac{397\sqrt{2}}{400}\)

- Direct substitution gives an indeterminate slope because both coordinate rates vanish. - Compare the lowest-order behavior of the rectangular coordinates near the cusp. - Use the limiting slope with the cusp point.

1. The rectangular coordinates are \(x=(1-\cos\theta)\cos\theta\) and \(y=(1-\cos\theta)\sin\theta\). 2. Both coordinate rates vanish at \(\theta=0\), so use the limiting slope rather than direct substitution. 3. Near \(0\), \(x\sim\frac{\theta^2}{2}\) and \(y\sim\frac{\theta^3}{2}\), so \(\frac{dy}{dx}\sim\frac{3\theta}{2}\to0\). 4. The cusp is at \((0,0)\), so the limiting tangent line is \(y=0\).

\(y=0\)