- Which listed rate describes vertical change, and which describes horizontal change? - Think of slope as vertical change per horizontal change. - Check that your result keeps the sign information from both rates.

1. Use the two listed component rates at the same parameter value. 2. Their quotient is \(\frac{dy}{dx}=\frac{5}{-3}=-\frac{5}{3}\).

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

- Compare the rate of change of \(y\) with the rate of change of \(x\). - What sign should a tangent slope have when \(x\) increases while \(y\) decreases? - Check that your result describes change in \(y\) per unit change in \(x\), not per unit change in \(t\).

1. The slope with respect to \(x\) is \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-6}{4}=-\frac{3}{2}\). 2. Because the slope is negative, the tangent falls as \(x\) increases.

\(\frac{dy}{dx}=-\frac{3}{2}\); the tangent falls as \(x\) increases.

- Which component graph is rising at the marked time? - Which component graph is falling? - Combine the signs of the two component rates to determine tangent slope.

1. In panel a), \(x(t)\) rises as \(t\) increases, so \(\frac{dx}{dt}>0\). 2. In panel b), \(y(t)\) falls as \(t\) increases, so \(\frac{dy}{dt}<0\). 3. Therefore, \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}<0\).

Negative: \(\frac{dy}{dt}<0\) and \(\frac{dx}{dt}>0\) at \(t=0\).

- A left-to-right tangent slope compares vertical change with horizontal change. - Check the signs of both given rates before dividing.

1. The tangent slope is the vertical rate divided by the horizontal rate. 2. \(\frac{dy}{dx}=\frac{-2}{4}=-\frac{1}{2}\). The tangent falls as \(x\) increases.

Falling; slope \(-\frac{1}{2}\).

- Picture a tiny increase in the parameter. How would the point move? - Use the two coordinate rates together to determine the tangent direction.

1. The horizontal coordinate has zero rate while the vertical coordinate is still changing. 2. The tangent is vertical, so its Cartesian slope is undefined.

Vertical tangent.

- Use the arrow to decide how each coordinate changes as the parameter increases. - The slope compares vertical change with horizontal change, not just the direction of travel.

1. The arrow points toward smaller \(x\)-values and smaller \(y\)-values, so both coordinates decrease. 2. A negative vertical change divided by a negative horizontal change is positive. The tangent slope is positive.

Both coordinates decrease; the tangent slope is positive.

- Imagine moving in the opposite direction along the same line. - What happens to the quotient if both its numerator and denominator change sign?

1. Reversing the direction changes both component rates to their negatives. 2. Their quotient is unchanged, so the tangent slope stays the same.

No. Both coordinate rates change sign, so their ratio stays the same.

- What two coordinate rates determine the direction of a parametric curve at one instant? - Find how each coordinate changes with \(t\) before using the requested parameter value. - Check that the horizontal component rate permits a finite tangent slope at \(t=1\).

1. Differentiate the components: \(\frac{dx}{dt}=2t + 2\) and \(\frac{dy}{dt}=3t^{2}-3\). 2. At \(t=1\), the component rates are \(\frac{dx}{dt}=4\) and \(\frac{dy}{dt}=0\). 3. Their quotient gives \(\frac{dy}{dx}=0\).

\(\frac{dy}{dx}=0\)

- Think about what the tangent slope should compare when both coordinates depend on the same parameter. - Differentiate each coordinate before evaluating anything at \(t=0\). - After finding the two local rates, check which one corresponds to horizontal change.

1. Differentiate the components: \(\frac{dx}{dt}=e^{t}+1\) and \(\frac{dy}{dt}=e^{t}-2\). 2. At \(t=0\), the component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=-1\). 3. Their quotient gives \(\frac{dy}{dx}=-\frac{1}{2}\).

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

- At the requested parameter, the tangent direction comes from how \(x\) and \(y\) are changing together. - Pay special attention to the signs when differentiating the trigonometric terms. - Use exact trigonometric values and then decide how the two component rates combine into slope.

1. Differentiate the components: \(\frac{dx}{dt}=\cos(t)+1\) and \(\frac{dy}{dt}=-\sin(t)-1\). 2. At \(t=\frac{\pi}{2}\), the component rates are \(\frac{dx}{dt}=1\) and \(\frac{dy}{dt}=-2\). 3. Their quotient gives \(\frac{dy}{dx}=-2\).

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

- Before differentiating, note the domain restriction created by the logarithm. - Work out the two coordinate rates carefully, especially the reciprocal term. - At \(t=1\), ask whether the horizontal rate is nonzero and what that means for a finite tangent slope.

1. Differentiate the components: \(\frac{dx}{dt}=2t + \frac{1}{t}\) and \(\frac{dy}{dt}=3 + \frac{1}{t^{2}}\). 2. At \(t=1\), the component rates are \(\frac{dx}{dt}=3\) and \(\frac{dy}{dt}=4\). 3. Their quotient gives \(\frac{dy}{dx}=\frac{4}{3}\).

\(\frac{dy}{dx}=\frac{4}{3}\)

- The tangent direction is determined by the two instantaneous coordinate changes, not by the coordinates themselves. - Differentiate the product in \(y\) before substituting the negative parameter value. - Once the two rates are known, check whether the horizontal rate allows an ordinary finite slope.

1. Differentiate the components: \(\frac{dx}{dt}=3t^{2}-1\) and \(\frac{dy}{dt}=2t+4\). 2. At \(t=-1\), the component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2\). 3. Their quotient gives \(\frac{dy}{dx}=1\).

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

- The given slope describes how the vertical and horizontal component rates compare. - Decide which component rate is unknown before writing a relationship among the three quantities. - Check whether the sign of your result is consistent with a negative tangent slope and a positive horizontal rate.

1. Relate the tangent slope to the two component rates. 2. Solve \(-\frac{1}{2}=\frac{dy/dt}{4}\) to obtain \(\frac{dy}{dt}=-2\).

\(\frac{dy}{dt}=-2\)

- Think about how tangent slope compares vertical and horizontal component rates. - Here the vertical rate is known, so determine what horizontal rate would make the stated slope possible. - Substitute your candidate rate back into the slope relationship as a check.

1. Relate the tangent slope to the two component rates. 2. Solve \(4=\frac{6}{dx/dt}\) to obtain \(\frac{dx}{dt}=\frac{3}{2}\).

\(\frac{dx}{dt}=\frac{3}{2}\)

- Compare the vertical and horizontal components of the same tangent direction. - Look for a common factor in the two rates. - Distinguish a vector’s magnitude from the direction it determines.

1. The tangent slope is the ratio of the vertical component rate to the horizontal component rate. 2. Therefore, \(\frac{dy}{dx}=\frac{-2q}{3q}=-\frac{2}{3}\). 3. The common nonzero factor \(q\) cancels, so changing \(q\) changes the magnitude of the component rates but not their direction ratio.

\(\frac{dy}{dx}=-\frac{2}{3}\); the common nonzero factor \(q\) cancels.

- How can you differentiate an integral whose upper limit is \(t\)? - Find the horizontal and vertical rates before forming the slope. - Check that the horizontal rate is nonzero at the requested value.

1. By the Fundamental Theorem of Calculus, \(\frac{dx}{dt}=1+t^2\) and \(\frac{dy}{dt}=3-t\). 2. At \(t=1\), both coordinate rates equal \(2\). 3. Thus \(\frac{dy}{dx}=\frac{2}{2}=1\).

\(1\)

- Use the two measured coordinate pairs, not the differences in \(t\) alone. - A secant through nearby points can estimate a tangent direction. - The estimate need not be the exact derivative.

1. The horizontal change between the points is \(0.2-(-0.2)=0.4\). 2. The vertical change is \(0.1-0.5=-0.4\). 3. The secant slope is \(\frac{-0.4}{0.4}=-1\), which estimates the nearby tangent slope.

Estimated tangent slope: \(-1\).

- What parameter value gives the marked point? - Find the two coordinate rates and decide which component, if any, is zero. - Match the resulting tangent direction to the labels on the graph.

1. Point P has \(x=0\), so it corresponds to \(t=0\). 2. Here \(\frac{dx}{dt}=1\) and \(\frac{dy}{dt}=2t\); at zero these are \(1\) and \(0\). 3. The tangent slope is \(0\), so the horizontal segment a lies along the tangent at P, unlike vertical segment b.

\(\frac{dx}{dt}=1\), \(\frac{dy}{dt}=0\); segment a.

- Read the line's rise over run from the grid, regardless of the arrow's direction. - Use the relationship between a tangent slope and the two coordinate rates.

1. From the grid, the line falls one unit when \(x\) increases one unit; its slope is \(-1\). 2. Since \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\), the vertical rate is \(-1\cdot 2=-2\).

\(\frac{dy}{dt}=-2\).

- What happens when you try to divide the two stated rates? - A zero vertical rate gives a horizontal tangent only if the horizontal rate is not zero.

1. The usual rate quotient is \(\frac{0}{0}\), which does not give a slope. 2. Different curves can still have different tangent directions at such points, so the rates alone are not enough.

No. Both rates are zero, so their quotient does not determine the tangent direction.

- Differentiate both coordinates without choosing a particular value of \(t\). - Do the derivatives depend on the parameter?

1. The coordinate rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=4\) at every parameter value. 2. The tangent slope is therefore always \(\frac{4}{2}=2\).

No. Its tangent slope is always \(2\).

- A tangent line needs both a point on the curve and its local direction there. - Use the parameter value separately in the coordinate functions and in the component rates. - Once the point and finite tangent slope are known, use any equivalent line form.

1. The point is \((5, 1)\). 2. Differentiate the components: \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so the tangent slope at \(t=2\) is \(\frac{dy}{dx}=2\). 3. Using point-slope form, an equation of the tangent line is \(y-1=2(x-5)\).

\(y-1=2(x-5)\)

- First distinguish the direction of the curve from the direction perpendicular to the curve. - Find the point and the tangent direction at the stated parameter before thinking about the normal line. - Check the geometric relationship between the tangent and normal slopes before writing the line.

1. The point is \((0, 9)\). 2. Differentiate the components: \(\frac{dx}{dt}=2t - 2\) and \(\frac{dy}{dt}=3t^{2}\), so the tangent slope at \(t=2\) is \(\frac{dy}{dx}=6\). 3. The normal slope is the negative reciprocal, \(-\frac{1}{6}\). 4. Using point-slope form, an equation of the normal line is \(y-9=-\frac{1}{6}x\).

\(y-9=-\frac{1}{6}x\)

- Separate the two pieces of information a tangent line needs: its contact point and its direction. - Keep the exact trigonometric values at \(t=\frac{\pi}{2}\) rather than converting them to decimals. - Check the signs of both component rates before deciding whether the tangent rises or falls as \(x\) increases.

1. The point is \(\left(0, 1 + \frac{\pi}{2}\right)\). 2. Differentiate the components: \(\frac{dx}{dt}=-\sin(t)\) and \(\frac{dy}{dt}=1 + \cos(t)\), so the tangent slope at \(t=\frac{\pi}{2}\) is \(\frac{dy}{dx}=-1\). 3. Using point-slope form, an equation of the tangent line is \(y-\left(1 + \frac{\pi}{2}\right)=-x\).

\(y-\left(1 + \frac{\pi}{2}\right)=-x\)

- The reciprocal terms affect the two component derivatives with opposite signs; check those signs carefully. - Determine the tangent direction at the parameter value before converting to the perpendicular direction. - Keep the fractional contact point exact when you write the normal line.

1. The point is \(\left(\frac{5}{2}, \frac{3}{2}\right)\). 2. Differentiate the components: \(\frac{dx}{dt}=1 - \frac{1}{t^{2}}\) and \(\frac{dy}{dt}=1 + \frac{1}{t^{2}}\), so the tangent slope at \(t=2\) is \(\frac{dy}{dx}=\frac{5}{3}\). 3. The normal slope is the negative reciprocal, \(-\frac{3}{5}\). 4. Using point-slope form, an equation of the normal line is \(y-\frac{3}{2}=-\frac{3}{5}\left(x-\frac{5}{2}\right)\).

\(y-\frac{3}{2}=-\frac{3}{5}\left(x-\frac{5}{2}\right)\)

- Find the contact point and the local direction as two separate pieces of the tangent-line problem. - The second coordinate is a product of two functions of \(t\); make sure its rate of change includes both contributions. - After evaluating at \(t=0\), check whether the tangent direction is finite before writing the line.

1. The point is \((1, 0)\). 2. Differentiate the components: \(\frac{dx}{dt}=e^{t}\) and \(\frac{dy}{dt}=e^{t}+te^{t}\), so the tangent slope at \(t=0\) is \(\frac{dy}{dx}=1\). 3. Using point-slope form, an equation of the tangent line is \(y=x-1\).

\(y=x-1\)

- What must be true about the curve's instantaneous vertical change for the tangent to be horizontal? - A candidate is not automatically valid; also consider what the horizontal coordinate is doing there. - After identifying the valid parameter values, return to the original coordinate equations to locate the points.

1. Differentiate: \(\frac{dx}{dt}=2t-4\) and \(\frac{dy}{dt}=3t^{2}-3\). 2. A horizontal tangent requires \(\frac{dy}{dt}=0\) and \(\frac{dx}{dt}\ne0\). 3. The valid parameter values are \(t=-1\) and \(t=1\). 4. Substitution gives the points \((5, 2)\) and \((-3, -2)\).

Parameter values: \(t=-1\), \(t=1\) Points: \((5, 2)\), \((-3, -2)\)

- Translate “horizontal tangent” into a statement about the two instantaneous coordinate changes. - Solve the relevant trigonometric condition over the entire interval, not just one quadrant. - Use the other component rate to screen the candidates before finding their coordinates.

1. Differentiate: \(\frac{dx}{dt}=-\sin(t)\) and \(\frac{dy}{dt}=2\cos(2t)\). 2. A horizontal tangent requires \(\frac{dy}{dt}=0\) and \(\frac{dx}{dt}\ne0\). 3. The valid parameter values are \(t=\frac{\pi}{4}\), \(t=\frac{3\pi}{4}\), \(t=\frac{5\pi}{4}\), and \(t=\frac{7\pi}{4}\). 4. Substitution gives the points \(\left(\frac{\sqrt{2}}{2}, 1\right)\), \(\left(-\frac{\sqrt{2}}{2}, -1\right)\), \(\left(-\frac{\sqrt{2}}{2}, 1\right)\), and \(\left(\frac{\sqrt{2}}{2}, -1\right)\).

Parameter values: \(t=\frac{\pi}{4}\), \(t=\frac{3\pi}{4}\), \(t=\frac{5\pi}{4}\), \(t=\frac{7\pi}{4}\) Points: \(\left(\frac{\sqrt{2}}{2}, 1\right)\), \(\left(-\frac{\sqrt{2}}{2}, -1\right)\), \(\left(-\frac{\sqrt{2}}{2}, 1\right)\), \(\left(\frac{\sqrt{2}}{2}, -1\right)\)

- For a vertical tangent, ask which coordinate must momentarily stop changing and which must still be changing. - Track all trigonometric solutions over \([0,2\pi]\); periodicity creates several candidates. - Verify the candidates before evaluating the original coordinates.

1. Differentiate: \(\frac{dx}{dt}=2\cos(2t)\) and \(\frac{dy}{dt}=-\sin(t)\). 2. A vertical tangent requires \(\frac{dx}{dt}=0\) and \(\frac{dy}{dt}\ne0\). 3. The valid parameter values are \(t=\frac{\pi}{4}\), \(t=\frac{3\pi}{4}\), \(t=\frac{5\pi}{4}\), and \(t=\frac{7\pi}{4}\). 4. Substitution gives the points \(\left(1, \frac{\sqrt{2}}{2}\right)\), \(\left(-1, -\frac{\sqrt{2}}{2}\right)\), \(\left(1, -\frac{\sqrt{2}}{2}\right)\), and \(\left(-1, \frac{\sqrt{2}}{2}\right)\).

Parameter values: \(t=\frac{\pi}{4}\), \(t=\frac{3\pi}{4}\), \(t=\frac{5\pi}{4}\), \(t=\frac{7\pi}{4}\) Points: \(\left(1, \frac{\sqrt{2}}{2}\right)\), \(\left(-1, -\frac{\sqrt{2}}{2}\right)\), \(\left(1, -\frac{\sqrt{2}}{2}\right)\), \(\left(-1, \frac{\sqrt{2}}{2}\right)\)

- The given tangent slope imposes a relationship between the two component rates at \(t=1\). - Differentiate while treating \(a\) as a constant, then evaluate the rates at the specified parameter. - After solving for \(a\), check that the horizontal component rate is not zero.

1. The component derivatives are \(\frac{dx}{dt}=a+2t\) and \(\frac{dy}{dt}=3t^{2}-1\). 2. At \(t=1\), set the quotient equal to \(2\): \(\frac{2}{a+2}=2\). 3. Solving gives \(a=-1\).

\(a=-1\)

- A slope of \(1\) means the instantaneous horizontal and vertical changes are related in a particularly simple way. - Find the component rates at \(t=0\) while keeping \(a\) unknown. - Check that the value you obtain produces an ordinary finite tangent slope.

1. The component derivatives are \(\frac{dx}{dt}=a\cos(t)+1\) and \(\frac{dy}{dt}=2-\sin(t)\). 2. At \(t=0\), set the quotient equal to \(1\): \(\frac{2}{a+1}=1\). 3. Solving gives \(a=1\).

\(a=1\)

- Keep the unknown coefficient in the horizontal rate instead of trying to solve from the original coordinates. - Evaluate the two component rates at \(t=0\) and use the stated geometric slope as a constraint between them. - Verify that your value of \(a\) leaves a nonzero horizontal rate.

1. The component derivatives are \(\frac{dx}{dt}=a+e^{t}\) and \(\frac{dy}{dt}=2e^{t}-1\). 2. At \(t=0\), set the quotient equal to \(\frac{1}{2}\): \(\frac{1}{a+1}=\frac{1}{2}\). 3. Solving gives \(a=1\).

\(a=1\)

- The unknown coefficient affects the local horizontal change, while the vertical change is determined once \(t=-1\) is used. - Differentiate first, then interpret the given negative slope as a relationship between the two rates. - Use the sign of the requested slope as a check on your final component rates.

1. The component derivatives are \(\frac{dx}{dt}=a+3t^{2}\) and \(\frac{dy}{dt}=2t-4\). 2. At \(t=-1\), set the quotient equal to \(-1\): \(-\frac{6}{a+3}=-1\). 3. Solving gives \(a=3\).

\(a=3\)

- Differentiate the two coordinates separately before checking the proposed slope. - A slope compares vertical change with horizontal change; which component belongs in the numerator? - Check where the horizontal rate is zero.

1. Differentiate: \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^2-2\). 2. Álvaro reversed the component rates. The slope is \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\), not \(\frac{dx/dt}{dy/dt}\). 3. Therefore, \(\frac{dy}{dx}=\frac{3t^2-2}{2t}=\frac{3t}{2}-\frac{1}{t}\) for \(t\ne0\).

The component derivatives are reversed. \(\frac{dy}{dx}=\frac{3t^2-2}{2t}=\frac{3t}{2}-\frac{1}{t}\) for \(t\ne0\).

- Compare the instantaneous horizontal and vertical rates, not the original coordinates. - Use exact trigonometric values at the stated angle. - The sign of the slope provides a quick check of the report.

1. The component rates are \(\frac{dx}{dt}=1+\cos t\) and \(\frac{dy}{dt}=-2\sin(2t)\). 2. At \(t=\frac{\pi}{4}\), these rates are \(1+\frac{\sqrt{2}}{2}\) and \(-2\). 3. Hence \(\frac{dy}{dx}=\frac{-2}{1+\frac{\sqrt{2}}{2}}=2\sqrt{2}-4\). This is negative, so it cannot equal the reported slope.

Incorrect. The tangent slope is \(2\sqrt{2}-4\).

- Find the horizontal and vertical component rates and compare them with the two proposed expressions. - Ordinary slope is vertical change per horizontal change; use that meaning to decide which expression has the correct orientation. - The simplified expression has a parameter restriction; identify where its denominator is zero.

1. Differentiate the coordinates: \(\frac{dx}{dt}=3t^{2}\) and \(\frac{dy}{dt}=2t\). 2. Form the quotient in the correct order: \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t}{3t^{2}}\). 3. Thus, \(\frac{dy}{dx}=\frac{2}{3t}\) for \(t\ne0\). The expression \(\frac{3t}{2}\) reverses the quotient.

\(\frac{dy}{dx}=\frac{2}{3t}\) for \(t\ne0\), because \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\). The other expression reverses the quotient.

- Find the horizontal and vertical rates separately, keeping track of the exponential sign. - Which rate should appear above the fraction bar in \(\frac{dy}{dx}\)? - Evaluate the corrected expression only after forming the quotient.

1. Differentiate: \(\frac{dx}{dt}=2e^{2t}\) and \(\frac{dy}{dt}=-e^{-t}\). 2. Jelena reversed the quotient. The correct derivative is \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=-\frac{e^{-3t}}{2}\). 3. At \(t=0\), the tangent slope is \(-\frac{1}{2}\).

Jelena reversed the quotient. The tangent slope is \(-\frac{1}{2}\).

- A parameter value must reproduce both coordinates of the given point, not just one of them. - If one coordinate equation gives more than one candidate, use the other coordinate to distinguish them. - After locating the correct parameter, use the local coordinate rates rather than the coordinates themselves to determine tangent slope.

1. From \(t^{2}-1=3\), the candidates are \(t=2\) and \(t=-2\). Only \(t=2\) also gives \(t^{3}-t=6\). 2. The derivatives are \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^{2}-1\). 3. At \(t=2\), \(\frac{dy}{dx}=\frac{11}{4}\).

\(t=2\); tangent slope: \(\frac{11}{4}\)

- Compare the given coordinates with both parametric coordinate equations. - Check whether one coordinate function is strictly monotonic to justify uniqueness. - Form the tangent slope from the component derivatives at the verified parameter value.

1. Substituting \(t=\pi\) gives \(x=2\pi - 1\) and \(y=\pi\), so \(t=\pi\) produces the point. 2. Because \(\frac{dx}{dt}=2-\sin(t)\ge 1\), the x-coordinate is strictly increasing, so no other parameter value produces the same point. 3. Also, \(\frac{dy}{dt}=1+\cos(t)\). 4. At \(t=\pi\), \(\frac{dy}{dx}=0\).

\(t=\pi\); tangent slope: \(0\)

- Solve one coordinate equation, then test every candidate in the other coordinate. - Differentiate the reciprocal terms with careful signs. - Evaluate the derivative quotient only at the verified parameter value.

1. The x-coordinate equation gives \(t=2\) or \(t=\frac{1}{2}\). Only \(t=2\) also gives \(y=\frac{3}{2}\). 2. The derivatives are \(\frac{dx}{dt}=1 - \frac{1}{t^{2}}\) and \(\frac{dy}{dt}=1 + \frac{1}{t^{2}}\). 3. At \(t=2\), \(\frac{dy}{dx}=\frac{5}{3}\).

\(t=2\); tangent slope: \(\frac{5}{3}\)

- Use one coordinate to recover the parameter and then check the other. - The tangent slope compares both coordinate rates at the same parameter. - Check that the horizontal rate is nonzero.

1. Since \(e^t=1\), the parameter is \(t=0\). This also gives \(x=0\). 2. The coordinate rates are \(\frac{dx}{dt}=1+\cos(t)\) and \(\frac{dy}{dt}=e^t\). 3. At \(t=0\), the rates are \(2\) and \(1\), so \(\frac{dy}{dx}=\frac{1}{2}\).

\(t=0\); tangent slope: \(\frac{1}{2}\)

- Solve the linear x-coordinate equation for \(t\) and substitute into the y-coordinate. - Differentiate the resulting Cartesian equation with respect to \(x\). - Compare that derivative with the parametric derivative at the point corresponding to \(t=2\).

1. From \(x=2t-1\), \(t=\frac{x+1}{2}\). Substituting into \(y=t^{2}+3\) gives \(y=\frac{x^{2}}{4}+\frac{x}{2}+\frac{13}{4}\). 2. Parametrically, \(\frac{dy}{dx}=\frac{2t}{2}=t\), so at \(t=2\) the slope is \(2\). 3. The corresponding point is \((3, 7)\). From the Cartesian equation, \(\frac{dy}{dx}=\frac{x}{2}+\frac{1}{2}\), which also gives \(2\) at \(x=3\).

The Cartesian equation is \(y=\frac{x^{2}}{4}+\frac{x}{2}+\frac{13}{4}\), and both methods give tangent slope \(2\).

- Multiply the two coordinate expressions to eliminate the exponential parameter. - Differentiate the Cartesian relation implicitly. - Evaluate both derivative methods at the point generated by \(t=0\).

1. Multiplying the coordinate equations gives \(xy=e^{t}e^{-t}=1\). 2. Parametrically, \(\frac{dy}{dx}=\frac{-e^{-t}}{e^{t}}=-e^{-2t}\), so at \(t=0\) the slope is \(-1\). 3. The corresponding point is \((1, 1)\). Implicit differentiation of \(xy=1\) gives \(y+x\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{y}{x}=-1\) at \((1,1)\).

The Cartesian relation is \(xy=1\), and both methods give tangent slope \(-1\) at \((1,1)\).

- Evaluate the signs of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) separately. - Use their quotient to decide whether \(y\) increases or decreases as \(x\) increases. - Use the two individual signs, not the quotient alone, to state the direction of motion.

1. The component rates are \(\frac{dx}{dt}=2t=1\) and \(\frac{dy}{dt}=3t^{2}-3=-\frac{9}{4}\). 2. The tangent slope is \(\frac{dy}{dx}=-\frac{9}{4}\), so the curve is locally decreasing as \(x\) increases. 3. Because \(\frac{dx}{dt}>0\) and \(\frac{dy}{dt}<0\), the motion for increasing \(t\) is to the right and downward.

Tangent slope: \(-\frac{9}{4}\); the curve is locally decreasing as \(x\) increases, and the increasing-\(t\) direction is right and downward.

- Differentiate both coordinates and evaluate the component rates at \(t=\pi\). - Use the quotient of the rates for the tangent slope. - Interpret the signs of the two rates to describe the parameter’s direction of motion.

1. The component rates are \(\frac{dx}{dt}=1-\sin(t)=1\) and \(\frac{dy}{dt}=-1+\cos(t)=-2\) at \(t=\pi\). 2. The tangent slope is \(\frac{dy}{dx}=-2\). 3. Because \(\frac{dx}{dt}>0\) and \(\frac{dy}{dt}<0\), the motion for increasing \(t\) is to the right and downward.

Tangent slope: \(-2\); the increasing-\(t\) direction is right and downward.

- Translate the word “parallel” into a condition on the tangent slope. - Express the tangent slope in terms of the parameter before solving on the stated interval. - Evaluate both coordinate functions after finding the valid parameter value.

1. The component derivatives are \(\frac{dx}{dt}=2t+2\) and \(\frac{dy}{dt}=3t^2-3\). 2. On the given interval, the tangent slope simplifies to \(\frac{dy}{dx}=\frac{3(t-1)}{2}\). 3. Parallel lines have equal slopes, so \(\frac{3(t-1)}{2}=3\), which gives \(t=3\). 4. At \(t=3\), the point is \((15,18)\).

\(t=3\), at the point \((15,18)\)

- First determine the point and tangent direction at the specified parameter value. - Use the tangent equation to locate where it meets each coordinate axis. - Treat the two intercept lengths as the base and height of a right triangle.

1. At \(t=1\), the curve passes through \((2,3)\). 2. The component derivatives are \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=-3t^2\), so the tangent slope is \(-\frac{3}{2}\). 3. The tangent line is \(y-3=-\frac{3}{2}(x-2)\), or \(y=-\frac{3}{2}x+6\). 4. The intercepts are \((4,0)\) and \((0,6)\), so the triangle’s area is \(\frac{1}{2}(4)(6)=12\).

\(12\) square units

- Express the point's x-coordinate and the tangent slope separately in terms of \(t\). - The slope comes from comparing the component rates. - After finding the parameter, check that the slope equals the x-coordinate.

1. The component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t-1\), so \(\frac{dy}{dx}=\frac{2t-1}{2}\). 2. Set the slope equal to \(x=2t+1\): \(\frac{2t-1}{2}=2t+1\), giving \(t=-\frac{3}{2}\). 3. The point is \(\left(-2,\frac{15}{4}\right)\). There, \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-4\), and \(\frac{dy}{dx}=-2\). 4. The tangent line is \(y-\frac{15}{4}=-2(x+2)\), or \(y=-2x-\frac{1}{4}\).

Point: \(\left(-2,\frac{15}{4}\right)\) Rates: \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-4\), \(\frac{dy}{dx}=-2\) Tangent: \(y=-2x-\frac{1}{4}\)

- Determine the secant direction from the two endpoint coordinates first. - Compute the component rates before forming the tangent slope. - The desired parameter must be strictly between the two endpoint parameter values.

1. The endpoints are \((2,2)\) and \((6,10)\), so the secant slope is \(\frac{10-2}{6-2}=2\). 2. The component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), giving tangent slope \(\frac{dy}{dx}=t\). 3. Parallelism requires \(t=2\), which lies between the endpoint parameter values. 4. The point is \((4,5)\), and there \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=4\), and \(\frac{dy}{dx}=2\).

Secant slope: \(2\) Required point: \((4,5)\) at \(t=2\) Rates: \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=4\), \(\frac{dy}{dx}=2\)

- Compare both component rates at the specified parameter value. - Think about the direction of a velocity vector whose horizontal component is zero. - A vertical line is written using a constant x-coordinate.

1. At \(t=1\), the point is \((-2,1)\). 2. The component rates are \(\frac{dx}{dt}=3t^2-3=0\) and \(\frac{dy}{dt}=2t=2\). 3. A zero horizontal rate with a nonzero vertical rate gives a vertical tangent, not a horizontal tangent. 4. The tangent-line equation is \(x=-2\).

Ji-hoon confused a zero horizontal rate with a zero slope. The tangent line is \(x=-2\).

- Find the point and tangent direction at the nearby known parameter value. - Treat the tangent line as a local model of the curve. - Use the small change in x-coordinate to estimate the corresponding change in y-coordinate.

1. At \(t=1\), the point is \((2,2)\). 2. The tangent slope is \(\frac{3t^2+1}{2t}=2\) at \(t=1\). 3. The tangent-line model is \(y-2=2(x-2)\). 4. At \(x=2.04\), the model gives \(y\approx2+2(0.04)=2.08\).

\(y\approx2.08\)

- First write the tangent slope in terms of the parameter. - Use the parameter restriction when solving the x-coordinate equation for \(t\). - Replace the parameter only after simplifying the derivative.

1. The parametric derivative is \(\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}\). 2. Since \(x=t^2+1\) and \(t>0\), \(t=\sqrt{x-1}\). 3. Substitution gives \(\frac{dy}{dx}=\frac{3}{2}\sqrt{x-1}\).

\(\frac{dy}{dx}=\frac{3}{2}\sqrt{x-1}\), for \(x>1\)

- Determine the tangent line before considering the distance. - Rewrite the line in a standard form with all terms on one side. - Use the coefficients of that line to measure the perpendicular distance from the origin.

1. At \(t=1\), the point is \((2,1)\), and the tangent slope is \(\frac{3}{2}\). 2. The tangent line is \(y-1=\frac{3}{2}(x-2)\), or \(3x-2y-4=0\). 3. The distance from \((0,0)\) to this line is \(\frac{|{-4}|}{\sqrt{3^2+(-2)^2}}=\frac{4}{\sqrt{13}}\).

\(\frac{4}{\sqrt{13}}\)

- First characterize the slope of any line whose two positive axis intercepts are equal. - Compare that slope with the ratio of the curve's component rates. - Verify the resulting line's two intercepts rather than assuming positivity.

1. A line with equal positive axis intercepts has slope \(-1\). 2. Here \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so \(\frac{dy}{dx}=t\). 3. Setting \(t=-1\) gives the point \((0,2)\), with component rates \(2\) and \(-2\). 4. The tangent line is \(y=-x+2\), whose two positive intercepts are both \(2\).

\(t=-1\) \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-2\), \(\frac{dy}{dx}=-1\) Tangent line: \(y=-x+2\)

- Check that both parameterized pieces approach the same point before comparing tangents. - Form the tangent slope from the component rates on each side of the join. - A single tangent requires the same limiting direction from both sides.

1. Both pieces approach \((0,0)\), so the curve is continuous at the origin. 2. From the left, \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=2t\), and \(\frac{dy}{dx}=t\to0\). 3. From the right, \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=2\), and \(\frac{dy}{dx}=1\). 4. The one-sided tangent slopes differ, so there is no single tangent line at the origin.

Left: \(\frac{dx}{dt}\to2\), \(\frac{dy}{dt}\to0\), \(\frac{dy}{dx}\to0\) Right: \(\frac{dx}{dt}\to2\), \(\frac{dy}{dt}\to2\), \(\frac{dy}{dx}\to1\) Therefore, no single tangent line exists at the origin.

- Differentiate each composed coordinate with respect to the parameter. - Match the given derivative values to the inputs produced at \(t=1\). - Form the tangent slope from the two resulting component rates.

1. The horizontal component rate is \(X'(t)=2t f'(t^2)\), so \(X'(1)=2(1)(4)=8\). 2. The vertical component rate is \(Y'(t)=3g'(3t-1)\), so \(Y'(1)=3(-3)=-9\). 3. Therefore, \(\frac{dY}{dX}=\frac{-9}{8}\).

\(-\frac{9}{8}\)

- Find the parameter value that produces the shared point on each curve before differentiating. - Compute each tangent direction from its own parameter-rate pair. - Sharing a point is not enough for tangency; compare the slopes before writing the common line.

1. The point \((2,1)\) corresponds to \(t=1\) on \(C_1\) and \(u=1\) on \(C_2\). 2. For \(C_1\), \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so at \(t=1\) the slope is \(1\). 3. For \(C_2\), \(\frac{dx}{du}=4u\) and \(\frac{dy}{du}=4\), so at \(u=1\) the slope is also \(1\). 4. The curves share the point and tangent slope, so the common tangent line is \(y-1=x-2\), or \(y=x-1\).

\(C_1\): \(t=1\), \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=2\), slope \(1\) \(C_2\): \(u=1\), \(\frac{dx}{du}=4\), \(\frac{dy}{du}=4\), slope \(1\) Common tangent: \(y=x-1\)

- Begin by finding where the horizontal coordinate has zero instantaneous rate. - If a candidate makes both coordinate rates zero, the usual quotient test is inconclusive; inspect the curve's local behavior there rather than classifying it immediately. - Only after resolving every candidate should you convert the valid parameter value to a point.

1. Differentiate: \(\frac{dx}{dt}=3t^{2}-3\) and \(\frac{dy}{dt}=2t+2\). 2. Solving \(\frac{dx}{dt}=0\) gives the candidates \(t=-1\) and \(t=1\). 3. At \(t=1\), \(\frac{dy}{dt}=4\ne0\), so the curve has a vertical tangent. 4. At \(t=-1\), both first derivatives are zero. Let \(u=t+1\). Then \(x=2-3u^{2}+u^{3}\), \(y=-1+u^{2}\), and \(\frac{dy}{dx}=\frac{2}{-6+3u}\to-\frac{1}{3}\), so the tangent there is not vertical. 5. Substituting \(t=1\) gives the point \((-2, 3)\).

Parameter value: \(t=1\) Point: \((-2, 3)\)

- Start with the condition suggested by the words “horizontal tangent,” but treat its solutions only as candidates. - If both coordinate rates vanish at one candidate, the ordinary slope quotient does not settle the tangent direction; investigate that point separately. - Keep the interval endpoints in your candidate list and verify every case before finding the points.

1. Differentiate: \(\frac{dx}{dt}=1+\cos(t)\) and \(\frac{dy}{dt}=-\sin(t)\). 2. Solving \(\frac{dy}{dt}=0\) gives the candidates \(t=0\), \(t=\pi\), and \(t=2\pi\). 3. At \(t=0\) and \(t=2\pi\), \(\frac{dx}{dt}=2\ne0\), so the tangents are horizontal. 4. At \(t=\pi\), both first derivatives are zero. Let \(u=t-\pi\). Then \(\frac{dy}{dx}=\frac{\sin(u)}{1-\cos(u)}=\cot\left(\frac{u}{2}\right)\), which is unbounded as \(u\to0\), so the tangent there is vertical rather than horizontal. 5. Substitution gives the points \((0, 1)\) and \((2\pi, 1)\).

Parameter values: \(t=0\), \(t=2\pi\) Points: \((0, 1)\), \((2\pi, 1)\)

- Identify every parameter value that produces the stated crossing point. - The curve can have a different tangent direction each time it visits the same point. - Keep the exact radical values when writing the two line equations.

1. Solving \(t^2=3\) gives the two parameter values \(t=-\sqrt{3}\) and \(t=\sqrt{3}\); both make \(y=0\). 2. The component derivatives are \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^2-3\). 3. At \(t=-\sqrt{3}\), the tangent slope is \(-\sqrt{3}\). At \(t=\sqrt{3}\), the tangent slope is \(\sqrt{3}\). 4. Using the common point \((3,0)\), the tangent lines are \(y=-\sqrt{3}(x-3)\) and \(y=\sqrt{3}(x-3)\).

\(y=-\sqrt{3}(x-3)\) and \(y=\sqrt{3}(x-3)\)

- Start with the two component rates; the tangent direction must come from their ratio. - Write a tangent line at a general parameter value before imposing the external-point condition. - Keep every parameter value that satisfies both the line condition and the original curve.

1. The component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so \(\frac{dy}{dx}=t\). 2. At parameter \(t\), the point is \((2t,t^2+1)\). Requiring its tangent line to pass through \((0,-3)\) gives \(-3-(t^2+1)=t(0-2t)\). 3. This simplifies to \(t^2=4\), so \(t=-2\) or \(t=2\). 4. At \(t=-2\), the rates are \(2\) and \(-4\), the slope is \(-2\), and the tangent is \(y=-2x-3\). 5. At \(t=2\), the rates are \(2\) and \(4\), the slope is \(2\), and the tangent is \(y=2x-3\).

\(t=-2\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-4\), \(\frac{dy}{dx}=-2\), \(y=-2x-3\) \(t=2\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=4\), \(\frac{dy}{dx}=2\), \(y=2x-3\)

- Convert perpendicularity into a relationship between the two line slopes. - Use the component rates to express the curve’s tangent slope in terms of \(t\). - Check that each candidate is in the interval and has a defined tangent slope.

1. The line \(4x+9y=12\) has slope \(-\frac{4}{9}\), so a perpendicular tangent must have slope \(\frac{9}{4}\). 2. The parametric tangent slope is \(\frac{dy}{dx}=\frac{3t^2-3}{2t}\), where \(t\ne0\). 3. Solving \(\frac{3t^2-3}{2t}=\frac{9}{4}\) gives \(2t^2-3t-2=0\), so \(t=-\frac{1}{2}\) or \(t=2\). 4. The corresponding points are \(\left(\frac{5}{4},\frac{11}{8}\right)\) and \((5,2)\).

\(t=-\frac{1}{2}\) at \(\left(\frac{5}{4},\frac{11}{8}\right)\) \(t=2\) at \((5,2)\)

- A tangent line must have both the correct direction and the correct point of contact. - Use the line’s slope to determine the unknown coefficient. - After finding the coefficient, check the curve’s point against the full line equation.

1. At \(t=1\), the tangent slope is \(\frac{3}{a+2}\). 2. Matching the slope of \(y=2x+1\) gives \(\frac{3}{a+2}=2\), so \(a=-\frac{1}{2}\). 3. The corresponding point is \(\left(\frac{1}{2},2\right)\). 4. Substitution into the stated line gives \(2=2\left(\frac{1}{2}\right)+1\), so the point lies on the line.

\(a=-\frac{1}{2}\); the point is \(\left(\frac{1}{2},2\right)\), which lies on \(y=2x+1\).

- Examine the tangent slope for nearby nonzero parameter values. - Simplify before taking the parameter toward the singular value. - Use the limiting direction together with the point at the singular value.

1. For \(t\ne0\), the tangent slope is \(\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}\). 2. As \(t\to0\), the tangent slope approaches \(0\). 3. The curve passes through \((0,0)\), so the limiting tangent line is \(y=0\).

\(y=0\)

- A specified tangent line gives one condition from its slope and another from the point of contact. - Express the point at the stated parameter value using both unknown constants. - Solve the two resulting conditions together.

1. At \(t=1\), the point is \((1+a,1+b)\). Because it lies on \(y=2x-1\), \(1+b=2(1+a)-1\), so \(b=2a\). 2. The tangent slope at \(t=1\) is \(\frac{3+b}{2}\). 3. Matching the stated line’s slope gives \(\frac{3+b}{2}=2\), so \(b=1\). 4. From \(b=2a\), \(a=\frac{1}{2}\).

\(a=\frac{1}{2}\) and \(b=1\)

- Write the tangent line at a general parameter value. - Express its y-intercept in terms of the parameter. - Use the given intercept to find the point of tangency.

1. For \(t\ne0\), the tangent slope is \(\frac{3t}{2}\). 2. The tangent line at parameter \(t\) has y-intercept \(t^3+1-\frac{3t}{2}(t^2)=1-\frac{t^3}{2}\). 3. Setting the intercept equal to \(-3\) gives \(1-\frac{t^3}{2}=-3\), so \(t=2\). 4. The point is \((4,9)\), the slope is \(3\), and the tangent line is \(y=3x-3\).

The point is \((4,9)\), and the tangent line is \(y=3x-3\).

- Build the tangent slope from the coordinate rates at the given parameter. - Substitute the parametric coordinates into the tangent-line equation to locate intersections. - The parameter at the tangency may be a repeated root.

1. At \(t=1\), \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=3\), and \(\frac{dy}{dx}=\frac{3}{2}\). The point is \((2,1)\). 2. The tangent is \(y-1=\frac{3}{2}(x-2)\), or \(y=\frac{3}{2}x-2\). 3. A second intersection satisfies \(t^3=\frac{3}{2}(2t)-2=3t-2\). 4. Factor \(t^3-3t+2=(t-1)^2(t+2)\). Besides the tangent's contact parameter, \(t=-2\) gives the point \((-4,-8)\).

At \(t=1\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=3\), \(\frac{dy}{dx}=\frac{3}{2}\) Tangent line: \(y=\frac{3}{2}x-2\) Other intersection: \((-4,-8)\)

- Convert the direction angle into a tangent-slope condition. - Solve the resulting equation only for positive parameter values. - Evaluate both coordinate functions for each valid solution.

1. A direction angle of \(\arctan(2)\) corresponds to tangent slope \(2\). 2. The parametric tangent slope is \(\frac{3t^2+1}{2t}\). 3. Solving \(\frac{3t^2+1}{2t}=2\) gives \(3t^2-4t+1=0\), so \(t=\frac{1}{3}\) or \(t=1\). 4. The corresponding points are \(\left(\frac{10}{9},\frac{10}{27}\right)\) and \((2,2)\).

\(\left(\frac{10}{9},\frac{10}{27}\right)\) and \((2,2)\)

- Treat the horizontal-tangent case separately: its normal is vertical. - For nonzero \(t\), find the tangent slope and then its negative reciprocal. - Impose the given external point on each normal-line equation.

1. The component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so the tangent slope is \(t\). 2. At \(t=0\), the tangent is horizontal, so the normal is the vertical line \(x=0\), which passes through \((0,3)\). 3. When \(t\ne0\), the normal slope is \(-\frac{1}{t}\). To pass through \((0,3)\), it must satisfy \(3-(t^2-1)=\left(-\frac{1}{t}\right)(0-2t)\). 4. This simplifies to \(4-t^2=2\), so \(t=\pm\sqrt{2}\). 5. The two other normals are \(y-3=-\frac{x}{\sqrt{2}}\) at \(t=\sqrt{2}\), and \(y-3=\frac{x}{\sqrt{2}}\) at \(t=-\sqrt{2}\).

\(t=0\): tangent slope \(0\); normal \(x=0\) \(t=\sqrt{2}\): tangent slope \(\sqrt{2}\); normal \(y-3=-\frac{x}{\sqrt{2}}\) \(t=-\sqrt{2}\): tangent slope \(-\sqrt{2}\); normal \(y-3=\frac{x}{\sqrt{2}}\)

- The two pieces must approach the same point at their join. - Find each one-sided tangent slope from the two coordinate rates. - Matching the geometric tangent does not require the two parameter speeds to match.

1. The first piece reaches \((2,2)\) at \(t=1\), so continuity requires \(a+b=2\). 2. The first piece has coordinate rates \(2t\) and \(3t^2\); its tangent slope at the join is \(\frac{3}{2}\). 3. The second piece has rates \(a\) and \(2t\), giving limiting tangent slope \(\frac{2}{a}\). Matching directions requires \(\frac{2}{a}=\frac{3}{2}\), hence \(a=\frac{4}{3}\). 4. Then \(b=2-\frac{4}{3}=\frac{2}{3}\).

\(a=\frac{4}{3}\) and \(b=\frac{2}{3}\)

- Write a tangent line using the curve's point and derivative at a general parameter. - The x-intercept is where the tangent's y-coordinate is zero. - Check any excluded parameter case before listing all solutions.

1. The coordinate rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), giving tangent slope \(t\). 2. For \(t\ne0\), the tangent at \((2t,t^2+1)\) has x-intercept \(2t-\frac{t^2+1}{t}=t-\frac{1}{t}\). At \(t=0\), the horizontal tangent has no x-intercept. 3. Solve \(t-\frac{1}{t}=\frac{3}{2}\): \(2t^2-3t-2=0\), so \(t=2\) or \(t=-\frac{1}{2}\). 4. At \(t=2\), the rates are \(2\) and \(4\), the slope is \(2\), and the tangent is \(y=2x-3\). 5. At \(t=-\frac{1}{2}\), the rates are \(2\) and \(-1\), the slope is \(-\frac{1}{2}\), and the tangent is \(y=-\frac{1}{2}x+\frac{3}{4}\).

\(t=2\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=4\), \(\frac{dy}{dx}=2\), \(y=2x-3\) \(t=-\frac{1}{2}\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-1\), \(\frac{dy}{dx}=-\frac{1}{2}\), \(y=-\frac{1}{2}x+\frac{3}{4}\)

- Find the contact point using the parameter and sign conditions. - Differentiate both equations with respect to \(t\), treating each coordinate as a function of \(t\). - Compare the two coordinate rates for the tangent slope.

1. At \(t=0\), the positive coordinate values are \(x=2\) and \(y=2\). 2. Differentiating \(x^2+\sin(t)=4\) gives \(2x\frac{dx}{dt}+\cos(t)=0\). Thus \(\frac{dx}{dt}=-\frac{1}{4}\) at \(t=0\). 3. Differentiating \(y^2+e^t=5\) gives \(2y\frac{dy}{dt}+e^t=0\). Thus \(\frac{dy}{dt}=-\frac{1}{4}\) at \(t=0\). 4. The tangent slope is \(1\), so the tangent is \(y-2=x-2\).

\(y-2=x-2\)

- Use the coordinate condition before working with the tangent slope. - Differentiate the two trigonometric coordinates with respect to the parameter. - Evaluate the remaining coordinate after finding both scale factors.

1. The x-coordinate condition gives \(a\cos\left(\frac{\pi}{3}\right)=3\), so \(a=6\). 2. The tangent slope is \(\frac{b\cos(t)}{-a\sin(t)}\). 3. At \(t=\frac{\pi}{3}\), the slope condition becomes \(-\frac{b}{6\sqrt{3}}=-\frac{1}{3}\), so \(b=2\sqrt{3}\). 4. The y-coordinate is \(2\sqrt{3}\sin\left(\frac{\pi}{3}\right)=3\), so the point is \((3,3)\).

\(a=6\), \(b=2\sqrt{3}\), and the point is \((3,3)\).

- The two curves have different parameters; find each tangent slope separately. - The angle calculation uses tangent slopes, not the parameter values. - Check that the angle you report is acute.

1. For \(C_1\), the coordinate rates at \(t=0\) are \(2\) and \(1\), so \(m_1=\frac{1}{2}\). 2. For \(C_2\), both coordinate rates at \(u=0\) equal \(1\), so \(m_2=1\). 3. The acute angle \(\phi\) satisfies \(\tan(\phi)=\left|\frac{m_2-m_1}{1+m_1m_2}\right|=\frac{1}{3}\). 4. Therefore, \(\phi=\arctan\left(\frac{1}{3}\right)\).

\(\arctan\left(\frac{1}{3}\right)\)

- Find the two coordinate rates for each curve at the shared parameter value. - A common tangent requires the same point and tangent direction. - Compare the signs of the vertical coordinates on either side of \(t=0\).

1. For \(C_1\), \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=3t^2\), and \(\frac{dy}{dx}=\frac{3}{2}t^2\), so its tangent slope at zero is \(0\). 2. For \(C_2\), \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=4t^3\), and \(\frac{dy}{dx}=2t^3\), giving the same tangent slope \(0\). 3. Both curves pass through the origin, so their common tangent is \(y=0\). 4. The quantity \(t^3\) changes sign across zero, so \(C_1\) crosses the tangent. Since \(t^4\ge0\) on both sides, \(C_2\) only touches it.

\(C_1\) at \(t=0\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=0\), \(\frac{dy}{dx}=0\) \(C_2\) at \(t=0\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=0\), \(\frac{dy}{dx}=0\) Common tangent: \(y=0\) \(C_1\) crosses it; \(C_2\) only touches it.