Free math worksheets

- Differentiate the x- and y-coordinate functions separately with respect to \(t\). - Form \(\frac{dy}{dx}\) as the quotient of the vertical and horizontal component rates. - Substitute \(t=1\) only after the derivative quotient is set up.

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

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

- Differentiate both the linear and exponential terms in each coordinate. - Put \(\frac{dy}{dt}\) in the numerator and \(\frac{dx}{dt}\) in the denominator. - Evaluate the exponential terms at \(t=0\) after forming the quotient.

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

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

- Differentiate the sine and cosine terms with their correct signs. - Build the parametric slope from the two component derivatives. - Use the exact trigonometric values at \(t=\frac{\pi}{2}\).

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

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

- Differentiate the logarithmic and reciprocal terms carefully, noting that \(t>0\). - Form the vertical-rate-to-horizontal-rate quotient before substituting. - Evaluate both component rates at \(t=1\) and verify the denominator is nonzero.

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

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

- Differentiate the cubic coordinate and the product in the second coordinate separately. - Use \(\frac{dy/dt}{dx/dt}\) for the tangent slope. - Substitute the negative parameter value carefully after differentiating.

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

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

- Use \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\). - Preserve the negative sign of the horizontal component rate. - Simplify the resulting fraction without converting it to a decimal.

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

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

- Start with the parametric slope formula and identify the missing numerator. - Multiply the known slope by the known horizontal component rate. - Check that the sign of the result is consistent with the given negative slope.

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

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

- Write the tangent slope as the vertical rate divided by the unknown horizontal rate. - Rearrange the equation before substituting any decimal approximations. - Verify the result by substituting it back into the quotient.

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.

- Differentiate each accumulated function with respect to its upper limit. - Evaluate the two component rates at the stated parameter value. - Compare the vertical rate with the horizontal rate.

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

\(1\)

- Evaluate both coordinate formulas at \(t=2\) to locate the point on the curve. - Since \(dx/dt\) is constant, divide the vertical rate \(2t\) by \(2\) at the stated parameter. - Insert the resulting point and slope into point-slope 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 \(y-1=2(x-5)\).

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

- Substitute \(t=2\) into the product and cubic coordinates before forming the line equation. - Compute the tangent slope from \((3t^2)/(2t-2)\) and verify the denominator is nonzero. - Take the negative reciprocal to obtain the normal slope through the point.

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

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

- Use the exact sine and cosine values at \(t=\frac{\pi}{2}\) to find the point. - Differentiate both trigonometric coordinates and form \((dy/dt)/(dx/dt)\). - The point has \(x=0\), which simplifies the point-slope equation.

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

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

- Evaluate the reciprocal terms at \(t=2\) to obtain the fractional point coordinates. - Differentiate \(1/t\) with the correct sign in each coordinate before forming the tangent slope. - Use the negative reciprocal of that slope for 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 \(y-\frac{3}{2}=-\frac{3}{5}\left(x-\frac{5}{2}\right)\).

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

- Evaluate \(e^t\) and \(te^t\) at \(t=0\) to find the point. - Apply the product rule to \(te^t\) before dividing by \(dx/dt=e^t\). - Use the resulting unit slope through \((1,0)\) to write the tangent 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 \(y=x-1\).

\(y=x-1\)

- Set the derivative of \(y=t^3-3t\) equal to zero to find horizontal-tangent candidates. - Check those candidates in \(dx/dt=2t-4\) so no vertical or singular case is included. - Evaluate both coordinate formulas at each surviving parameter value.

1. Differentiate: \(\frac{dx}{dt}=2t - 4\) and \(\frac{dy}{dt}=3t^{2} - 3\). 2. A horizontal tangent requires \(\frac{dy}{dt}=0\) and \(\frac{dx}{dt}\ne 0\). 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: \((5, 2)\), \((-3, -2)\)

- Solve for the parameter values where the horizontal component rate is zero. - If both component rates vanish at a candidate, examine the limiting slope rather than discarding it automatically. - Substitute only the valid parameter values into both coordinate functions.

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 \(t=1\), \(\frac{dy}{dt}=4\ne 0\), 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: \((-2, 3)\)

- Solve \(\cos(2t)=0\) over the full interval \([0,2\pi]\). - Verify that \(-\sin t\) is nonzero at every candidate before calling the tangent horizontal. - Use exact unit-circle values for \(\cos t\) and \(\sin(2t)\) to list the points.

1. Differentiate: \(\frac{dx}{dt}=-\sin(t)\) and \(\frac{dy}{dt}=2\cos(2t)\). 2. A horizontal tangent requires \(\frac{dy}{dt}=0\) and \(\frac{dx}{dt}\ne 0\). 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: \(\left(\frac{\sqrt{2}}{2}, 1\right)\), \(\left(-\frac{\sqrt{2}}{2}, -1\right)\), \(\left(-\frac{\sqrt{2}}{2}, 1\right)\), \(\left(\frac{\sqrt{2}}{2}, -1\right)\)

- Find where \(2\cos(2t)\) vanishes on \([0,2\pi]\). - Check that \(-\sin t\) does not vanish at the same parameters. - Evaluate \(\sin(2t)\) and \(\cos t\) exactly to pair each parameter with its point.

1. Differentiate: \(\frac{dx}{dt}=2\cos(2t)\) and \(\frac{dy}{dt}=-\sin(t)\). 2. A vertical tangent requires \(\frac{dx}{dt}=0\) and \(\frac{dy}{dt}\ne 0\). 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: \(\left(1, \frac{\sqrt{2}}{2}\right)\), \(\left(-1, -\frac{\sqrt{2}}{2}\right)\), \(\left(1, -\frac{\sqrt{2}}{2}\right)\), \(\left(-1, \frac{\sqrt{2}}{2}\right)\)

- Solve for the parameter values where the vertical component rate is zero. - If both component rates vanish at a candidate, examine the limiting slope instead of discarding it automatically. - Substitute only the parameter values that actually produce horizontal tangents.

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 \(t=2\pi\), \(\frac{dx}{dt}=2\ne 0\), 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 \(u\to 0\), 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: \((0, 1)\), \((2\pi, 1)\)

- Differentiate both coordinate functions while treating \(a\) as a constant. - Evaluate the component derivatives at \(t=1\) and set their quotient equal to the stated slope. - Verify that the resulting horizontal component derivative is nonzero.

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 \(a=-1\).

\(a=-1\)

- Differentiate the trigonometric and linear terms in both coordinates. - Substitute \(t=0\) into the component derivatives before forming the slope equation. - Solve the resulting equation for \(a\) and check the denominator.

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 \(a=1\).

\(a=1\)

- Differentiate the exponential terms carefully in both coordinates. - Use the specified slope to relate \(\frac{dy}{dt}\) and \(\frac{dx}{dt}\) at \(t=0\). - Confirm that the candidate value does not make \(\frac{dx}{dt}=0\).

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 \(a=1\).

\(a=1\)

- Differentiate the product in the x-coordinate and the quadratic expression in the y-coordinate. - Evaluate at \(t=-1\) before setting the derivative quotient equal to the given slope. - Check the sign of each component derivative when solving for \(a\).

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 each coordinate independently before examining the proposed quotient. - Recall that the vertical component derivative belongs in the numerator. - State where the corrected quotient is undefined because the horizontal component rate vanishes.

1. Differentiate the coordinates: \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^{2} - 2\). 2. The student reversed the component derivatives. The tangent derivative 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 \(t\ne 0\).

The student reversed the component derivatives. The correct derivative is \(\frac{dy}{dx}=\frac{3t^{2} - 2}{2t}=\frac{3t}{2} - \frac{1}{t}\), for \(t\ne 0\).

- Use the chain rule when differentiating the double-angle cosine term. - Form \(\frac{dy/dt}{dx/dt}\) before substituting the parameter value. - Compare the exact resulting slope with the student’s claimed value.

1. Differentiate the coordinates: \(\frac{dx}{dt}=\cos(t)\) and \(\frac{dy}{dt}=-2\sin(2t)\). 2. Therefore, \(\frac{dy}{dx}=\frac{-2\sin(2t)}{\cos(t)}=-4\sin(t)\) wherever \(\cos(t)\ne0\). 3. At \(t=\frac{\pi}{4}\), the slope is \(-2\sqrt{2}\), not \(2\).

The report is incorrect. At \(t=\frac{\pi}{4}\), the tangent slope is \(-2\sqrt{2}\).

- Differentiate the two power functions independently. - Place \(\frac{dy}{dt}\) over \(\frac{dx}{dt}\), then simplify common factors. - Include the parameter restriction created by the denominator.

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.

- Apply the chain rule to both exponential coordinate functions. - Use the vertical derivative divided by the horizontal derivative. - Evaluate the corrected quotient at the specified parameter value.

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

The student reversed the quotient. The tangent slope at \(t=0\) is \(-\frac{1}{2}\).

- Solve the x-coordinate equation for all possible parameter values. - Use the y-coordinate to determine which candidate actually produces the given point. - Evaluate the parametric derivative quotient at that parameter value.

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 the exponential x-coordinate to identify the parameter value directly. - Confirm that the same value produces the stated y-coordinate. - Differentiate both exponentials and form the tangent-slope quotient.

1. Since \(e^{t}=1\), the parameter value is \(t=0\); this also gives \(e^{2t}=1\). 2. The derivatives are \(\frac{dx}{dt}=e^{t}\) and \(\frac{dy}{dt}=2e^{2t}\). 3. At \(t=0\), \(\frac{dy}{dx}=2\).

\(t=0\); tangent slope: \(2\)

- 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 both the tangent slope and the requested coordinate in terms of the parameter. - Use the stated equality to determine the parameter value. - Substitute that value into both coordinate functions before writing the line.

1. The tangent slope is \(\frac{dy}{dx}=2t-1\), and the x-coordinate is \(t+1\). 2. Setting these equal gives \(2t-1=t+1\), so \(t=2\). 3. The point is \((3,2)\), and the tangent slope is \(3\). 4. The tangent line is \(y-2=3(x-3)\).

The point is \((3,2)\), and the tangent line is \(y-2=3(x-3)\).

- First determine the direction of the line through the two endpoint points. - Express the tangent direction at a general parameter value. - Solve for the parameter between the endpoints, then evaluate the coordinates.

1. The endpoint points are \((1,2)\) and \((3,10)\), so the secant slope is \(\frac{10-2}{3-1}=4\). 2. The tangent slope is \(\frac{dy}{dx}=2t\). 3. Setting the tangent slope equal to the secant slope gives \(2t=4\), so \(t=2\). 4. The corresponding point is \((2,5)\).

\((2,5)\)

- 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 \(x=-2\).

The student 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 \(\frac{|{-4}|}{\sqrt{3^2+(-2)^2}}=\frac{4}{\sqrt{13}}\).

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

- Think about the slope of a line that joins equal positive intercepts on the two axes. - Match that slope to the curve’s tangent direction. - Verify the intercepts after writing the tangent equation.

1. A line with equal positive axis intercepts has slope \(-1\). 2. The tangent slope is \(\frac{dy}{dx}=2t\), so \(2t=-1\) gives \(t=-\frac{1}{2}\). 3. The point is \(\left(\frac{3}{2},\frac{5}{4}\right)\). 4. The tangent line is \(y-\frac{5}{4}=-(x-\frac{3}{2})\), or \(x+y=\frac{11}{4}\). Both intercepts are \(\frac{11}{4}\).

\(x+y=\frac{11}{4}\)

- First check that both pieces meet at the same point. - Find the tangent direction on each side of the joining parameter. - A single tangent requires the two limiting directions to agree.

1. Both pieces approach \((0,0)\), so the curve is continuous at the origin. 2. From the left, the tangent slope is \(2t\), which approaches \(0\). 3. From the right, the tangent slope is constantly \(2\). 4. Because the one-sided tangent slopes differ, the curve has a corner and no single tangent line at the origin.

No. The left-hand slope is \(0\) and the right-hand slope is \(2\), so no single tangent line exists.

- 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, \(\frac{dY}{dX}=\frac{-9}{8}\).

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

- Use the parameter value and sign information to identify the point. - Differentiate each coordinate relation with respect to the parameter. - Combine the two component rates to obtain the tangent direction.

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

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

- Verify the parameter value for the shared point on each curve. - Compute the two tangent directions independently. - Curves are tangent when they share both a point and a tangent line.

1. The point \((1,1)\) corresponds to \(t=1\) on \(C_1\) and \(u=1\) on \(C_2\). 2. For \(C_1\), the tangent slope is \(2t=2\) at \(t=1\). 3. For \(C_2\), the tangent slope is \(\frac{4}{2u}=2\) at \(u=1\). 4. The curves share both a point and a tangent slope, so they are tangent there. Their common tangent line is \(y-1=2(x-1)\).

Both tangent slopes equal \(2\), and the common tangent line is \(y-1=2(x-1)\).

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

- Write the tangent line at a general parameter value before using the external point. - Impose the condition that the given point lies on that line. - Check every resulting parameter value in its corresponding tangent equation.

1. At parameter \(t\), the point is \((t,t^2+1)\) and the tangent slope is \(2t\). 2. Requiring the tangent line through \((t,t^2+1)\) to contain \((0,-3)\) gives \(-3-(t^2+1)=2t(0-t)\). 3. The condition simplifies to \(t^2=4\), so \(t=-2\) or \(t=2\). 4. At \(t=-2\), the tangent line is \(y=-4x-3\). At \(t=2\), the tangent line is \(y=4x-3\).

\(t=-2\): \(y=-4x-3\) \(t=2\): \(y=4x-3\)

- 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 \(y=3x-3\).

- First find the tangent line at the stated parameter value. - A second intersection must satisfy both the curve equations and the line equation. - The repeated root corresponds to tangency; look for any remaining root.

1. At \(t=1\), the point is \((1,1)\) and the tangent slope is \(3\). 2. The tangent line is \(y-1=3(x-1)\), or \(y=3x-2\). 3. On the curve, intersection with the line requires \(t^3=3t-2\). 4. Factoring gives \((t-1)^2(t+2)=0\). The other parameter value is \(t=-2\), which gives \((-2,-8)\).

\((-2,-8)\)

- 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 \((2,2)\)

- Check separately whether the parameter value that gives a horizontal tangent produces a vertical normal through the origin. - For nonzero parameter values, express the normal slope as the negative reciprocal of the tangent slope. - Impose the condition that the origin lies on each normal line.

1. At \(t=0\), the tangent is horizontal, so the normal line is vertical. Because the point is \((0,-1)\), this normal line is \(x=0\), which passes through the origin. 2. For \(t\ne0\), the tangent slope is \(2t\), so the normal slope is \(-\frac{1}{2t}\). 3. Requiring the normal through \((t,t^2-1)\) to contain the origin gives \(t^2-1=-\frac{1}{2t}(t)=-\frac{1}{2}\). 4. Thus, \(t^2=\frac{1}{2}\), so \(t=\pm\frac{1}{\sqrt{2}}\). 5. The corresponding normal lines are \(y=-\frac{1}{\sqrt{2}}x\) and \(y=\frac{1}{\sqrt{2}}x\).

\(x=0\), \(y=-\frac{1}{\sqrt{2}}x\), and \(y=\frac{1}{\sqrt{2}}x\)

- Continuity and tangent agreement impose separate conditions. - Match the two coordinate values at the joining parameter. - Compare the one-sided tangent slopes after enforcing continuity.

1. The left-hand piece reaches \((1,1)\) at \(t=1\). Continuity requires \(a+b=1\). 2. The left-hand tangent slope at \(t=1\) is \(2\). 3. The right-hand tangent slope is \(\frac{2}{a}\), so matching slopes gives \(\frac{2}{a}=2\), hence \(a=1\). 4. From \(a+b=1\), \(b=0\).

\(a=1\) and \(b=0\)

- Write the tangent line at a general nonzero parameter value. - Determine where that line meets the x-axis. - Solve the intercept condition and write a line for each valid parameter.

1. At parameter \(t\ne0\), the point is \((t,t^2+1)\) and the tangent slope is \(2t\). 2. The tangent line’s x-intercept is \(t-\frac{t^2+1}{2t}=\frac{t^2-1}{2t}\). 3. Setting this equal to \(\frac{3}{4}\) gives \(2t^2-3t-2=0\), so \(t=2\) or \(t=-\frac{1}{2}\). 4. The tangent lines are \(y=4x-3\) and \(y=-x+\frac{3}{4}\).

\(y=4x-3\) and \(y=-x+\frac{3}{4}\)

- 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 \((3,3)\).

- Find the parameter value on each curve that produces the intersection point. - Determine the two tangent slopes independently. - Use a relationship that converts two line slopes into the acute angle between them.

1. The point \((1,1)\) corresponds to \(t=1\) on \(C_1\) and \(u=1\) on \(C_2\). 2. The tangent slopes are \(m_1=2\) and \(m_2=\frac{1}{2}\). 3. The acute angle \(\phi\) satisfies \(\tan(\phi)=\left|\frac{m_1-m_2}{1+m_1m_2}\right|=\frac{3}{4}\). 4. Therefore, \(\phi=\arctan\left(\frac{3}{4}\right)\).

\(\arctan\left(\frac{3}{4}\right)\)

- Determine the tangent direction of each curve at the shared point. - After finding the common line, inspect the sign of each y-coordinate near the parameter value. - Crossing requires the curve to lie on opposite sides of the tangent line.

1. For both curves, \(\frac{dx}{dt}=1\). Their vertical derivatives at \(t=0\) are both \(0\), so each tangent slope is \(0\). 2. The common tangent line is \(y=0\). 3. For \(C_1\), \(y=t^3\) changes sign as \(t\) passes through \(0\), so the curve crosses the tangent line. 4. For \(C_2\), \(y=t^4\ge0\) on both sides of \(0\), so the curve touches the tangent line without crossing it.

The common tangent line is \(y=0\). Curve \(C_1\) crosses it, while curve \(C_2\) only touches it.