Parametric equations-feature image

Parametric Equations

Parametric Equations

The traditional representation of y as a function of x or \({\small y = f(x)}\) is inadequate to represent a curve or a surface.

To be able to draw a curve or a surface, we need to separate the x and y and write them in terms of an independent variable.

Parametric equations are used to express the Cartesian coordinates (x and y) in terms of another independent variable, usually named as t.

The typical procedure in this topic is to find the gradient of the curve or \({ \large\frac{\mathrm{d}y}{\mathrm{d}x} }\). We can do this by finding each derivative of x and y with respect to t and then divide them both.

Let’s dig into some of the examples to show you what I mean. Cheers ! =) .
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EXAMPLE:

\({\small 1.\enspace}\) The parametric equations of a curve are
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\(\hspace{3em} x = {\large \frac{1}{{\cos}^{3}t}}, \quad y = {\tan}^{3}t\).
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where \(0 \ \le \ t \ \le \ \frac{\pi}{2}\)
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\({\small\hspace{1.2em}\left(a\right). \enspace }\) Show that \( {\large\frac{\mathrm{d}y}{\mathrm{d}x} } = \sin \ t\)
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\({\small\hspace{1.2em}\left(b\right). \enspace }\) Hence, show that the equation of the tangent to the curve at the point with parameter t is \( y = x \sin t \ – \ \tan t\).

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\({\small 2.\enspace}\) The parametric equations of a curve are
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\(\hspace{3em} x = e^{-t} \cos t, \quad y = e^{-t} \sin t\).
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Show that \( {\large\frac{\mathrm{d}y}{\mathrm{d}x} } = \tan \big(t \ – \ {\large\frac{\pi}{4}}\big)\)

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\({\small 3.\enspace}\) The parametric equations of a curve are
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\(\hspace{3em} x = \ln (2t \ + \ 3), \quad y = { \large\frac{3t \ + \ 2}{2t \ + \ 3} }\).
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Find the gradient of the curve at the point where it crosses the y-axis.

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\({\small 4.\enspace}\) The parametric equations of a curve are
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\( {\small x = \ 2\sin \theta \ + \ \sin 2\theta, \enspace y = \ 2\cos \theta \ + \ \cos 2\theta }\),
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where \(0 \ \lt \ \theta \ \lt \ \pi\)
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\({\small\hspace{1.2em}\left(\textrm{i}\right). \enspace }\) Obtain an expression for \( {\small {\large\frac{\mathrm{d}y}{\mathrm{d}x} } }\) in terms of \({\small \theta }\).
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\({\small\hspace{1.2em}\left(\textrm{ii}\right). \enspace }\) Hence find the exact coordinates of the point on the curve at which the tangent is parallel to the y-axis.

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\({\small 5.\enspace}\) The parametric equations of a curve are
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\( x = \ 2 t \ + \sin 2t, \enspace y = \ 1 \ – \ 2\cos 2t \),
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where \(-\frac{1}{2}\pi \ \lt \ t \ \lt \ \frac{1}{2}\pi\)
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\({\small\hspace{1.2em}\left(\textrm{i}\right). \enspace }\) Show that \( {\small {\large\frac{\mathrm{d}y}{\mathrm{d}x} } \ = \ 2 \tan t }\).
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\({\small\hspace{1.2em}\left(\textrm{ii}\right). \enspace }\) Hence find the x-coordinate of the point on the curve at which the gradient of the normal is 2. Give your answer correct to 3 significant figures.

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PRACTICE MORE WITH THESE QUESTIONS BELOW!

\({\small 1.\enspace}\) The parametric equations of a curve are

\( x = \ \sin t \ + \cos t, \enspace y = \ {\sin}^{3}t \ + \ {\cos}^{3}t\),

where \(\frac{\pi}{4} \ \le \ t \ \le \ \frac{5\pi}{4}\)

\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Show that \({ \large\frac{\mathrm{d}y}{\mathrm{d}x}} = \ -3 \ \sin t \ \cos t \).
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Find the gradient of the curve at the origin.
\({\small\hspace{1.2em}\left(c\right).\hspace{0.8em}}\) Find the values of t for which the gradient of the curve is 1, giving your answers correct to 2 significant figures.

\({\small 2. \enspace}\) The parametric equations of a curve are

\( x = \ a(2\theta \ – \ \sin 2\theta), \enspace y = \ a(1 \ – \ \cos 2\theta)\).

Show that \({ \large\frac{\mathrm{d}y}{\mathrm{d}x}} = \ \cot \theta \).

\({\small 3.\enspace}\) The parametric equations of a curve are

\( x = \ \ln \cos \theta, \enspace y = \ 3\theta \ – \ \tan \theta\),

where \(0 \ \le \ \theta \ \le \ \frac{1}{2}\pi\).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Express \({\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} }\) in terms of \({\small \tan \theta}\).
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) Find the exact y-coordinate of the point on the curve at which the gradient of the normal is equal to 1.

\({\small 4.\enspace}\) The parametric equations of a curve are

\( x = \ {t}^{2} \ + 1, \enspace y = \ 4t \ + \ \ln \ (2t \ – \ 1)\).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Express \({\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} }\) in terms of \({\small t}\).
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) Find the equation of the normal to the curve at the point where \({\small t \ = \ 1}\). Give your answer in the form ax + by + cz = 0.

\({\small 5.\enspace}\) The parametric equations of a curve are

\( x = \ t \ + \cos t, \enspace y = \ \ln \ (1 + \sin t)\),

where \(-\frac{1}{2}\pi \ \lt \ t \ \lt \ \frac{1}{2}\pi\).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Show that \({\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ \sec t}\).
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) Hence find the x-coordinates of the points on the curve at which the gradient is equal to 3. Give your answer correct to 3 significant figures.

\({\small 6.\enspace}\) A curve has parametric equations

\( x = \ {t}^{2} \ + \ 3t \ + \ 1, \enspace y = \ {t}^{4} \ + \ 1\).

The point P on the curve has parameter p. It is given that the gradient of the curve at P is 4. Show that \({\small p \ = \ \sqrt[3]{(2p \ + \ 3)} }\).

\({\small 7.\enspace}\) A curve has parametric equations

\( x = \ 1 \ + \ 2 \ \sin \theta \ \) and \( \ y = \ 4 \ + \ \sqrt{3} \ \cos \theta\).

\({\small\hspace{1.2em}\left(a\right).\hspace{0.8em}}\) Find the equations of the tangent and normal at the point P where \({\small \theta \ = \ {\large\frac{\pi}{6}} }\). Hence, find the area A of the triangle bounded by the tangent and normal at P, and the y-axis.
\({\small\hspace{1.2em}\left(b\right).\hspace{0.8em}}\) Determine the rate of change of xy at \({\small \theta \ = \ {\large\frac{\pi}{6}} }\) if x increases at a constant rate of 0.1 units/s.

\({\small 8.\enspace}\) A curve is defined parametrically by \( x \ = \ \frac{2t}{t + 1} \) and \( y \ = \ \frac{{t}^{2}}{t + 1} \).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Find the equation of the normal to the curve at the point P(1, 1/2).
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) The normal at P meets the curve again at Q. Find the exact coordinates of Q.

\({\small 9.\enspace}\) A curve has parametric equations:

\( x = \ \sec (\frac{\theta}{2}), \enspace y = \ \ln \ \sec (\frac{\theta}{2})\),

where \(-\pi \ \lt \ \theta \ \lt \ \pi \).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Find the equation of the normal to the curve at the point at \({\small \theta \ = \ {\large\frac{\pi}{3}} }\).
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) Determine the rate of change of x if the gradient of the curve at \({\small \theta \ = \ {\large\frac{\pi}{2}} }\) is decreasing at a rate of 0.4 units per second.

\({\small 10.\enspace}\) The parametric equations of a curve are

\( x = \ t \ + \ \ln t, \enspace y = \ t \ + \ {\textrm{e}}^{t}\) for \( t \gt 0\).

\({\small\hspace{1.2em}\left(\textrm{i}\right).\hspace{0.8em}}\) Sketch the curve, indicating clearly all intercepts and asymptotes.
\({\small\hspace{1.2em}\left(\textrm{ii}\right).\hspace{0.6em}}\) Show that, for all the points on the curve, \({\small { \large\frac{\mathrm{d}y}{\mathrm{d}x}} \ = \ { \large\frac{t(1 \ + \ {\textrm{e}}^{t})}{t \ + \ 1}} }\).
Hence, deduce that the curve does not have any turning points.
\({\small\hspace{1.2em}\left(\textrm{iii}\right).\hspace{0.4em}}\) Find, in exact form, the equation of the normal of the curve at the point where \({\small t \ = \ 1}\).


As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .

4 comments
  1. Question2

  2. Hi sir, how do you solve for question 9ii, the one with x=sec(theta/2) and y=lnsec(theta/2)? thanks!

  3. What is the solution for question 3?

    1. Hi Tallulah,
      You can find \({\large\frac{\mathrm{d}y}{\mathrm{d}x} } = {\large\frac{\mathrm{d}y}{\mathrm{d}\theta} } \ \div \ {\large\frac{\mathrm{d}x}{\mathrm{d}\theta} }\).

      And then, for part (b), you can find the gradient of the tangent of the curve by using the perpendicular gradient formula, \({\small m_{1} \ \times \ m_{2} \ = \ -1 }\).
      Since \({\large\frac{\mathrm{d}y}{\mathrm{d}x} } \) is the gradient of the tangent of the curve, you can equate them to find the x value and substitute the x value to get the y-coordinate of the point.

      Cheers,
      Mr Will

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