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Q2RE

Expert-verifiedFound in: Page 535

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**(a) How do you find the slope of a tangent to a parametric curve? **

**(b) How do you find the area under a parametric curve?**

- For a parametric curve \({\rm{x = f(t), y = g(t),}}\) the slope of the tangent is \(\frac{{{\rm{g'(t)}}}}{{{\rm{f'(t)}}}}\).
- For a parametric curve \({\rm{x = f(t), y = g(t),}}\)the area under the curve is \({\rm{A = }}\int_{\rm{\alpha }}^{\rm{\beta }} {{\rm{g(t)f'(t)dt}}} \).

**Parametric curves allow us to draw relationships between two or more numbers while also representing the directions or orientations of each quantity.**

(a)

The parametric curve is represented as –

\({\rm{x = f(t), y = g(t)}}\)

Differentiate the functions with respect to \({\rm{t}}\)–

\(\frac{{{\rm{dx}}}}{{{\rm{dt}}}}{\rm{ = f'(t), }}\frac{{{\rm{dy}}}}{{{\rm{dt}}}}{\rm{ = g'(t)}}\)

Now use the chain rule to obtain the slope –

Slope of tangent\({\rm{ = }}\frac{{{\rm{dy}}}}{{{\rm{dx}}}}\)

\(\begin{aligned}{c}{\rm{ = }}\frac{{{\rm{dy}}}}{{{\rm{dt}}}} \cdot \frac{{{\rm{dx}}}}{{{\rm{dt}}}}\\{\rm{ = }}\frac{{{{{\rm{dy}}} \mathord{\left/

{\vphantom {{{\rm{dy}}} {{\rm{dt}}}}} \right.

\kern-\nulldelimiterspace} {{\rm{dt}}}}}}{{{{{\rm{dx}}} \mathord{\left/

{\vphantom {{{\rm{dx}}} {{\rm{dt}}}}} \right.

\kern-\nulldelimiterspace} {{\rm{dt}}}}}}{\rm{ = }}\frac{{{\rm{g'(t)}}}}{{{\rm{f'(t)}}}}\end{aligned}\)

Therefore, the result is obtained as \(\frac{{{\rm{g'(t)}}}}{{{\rm{f'(t)}}}}\).

(b)

The area under the curve \({\rm{y = F(x)}}\) from \({\rm{a}}\) to \({\rm{b}}\) is given by \({\rm{A = }}\int_{\rm{a}}^{\rm{b}} {{\rm{F(x)dx}}} \) where \({\rm{F(x)}} \ge {\rm{0}}\).

If curve is given by parametric equations –

\({\rm{x = f(t), y = g(t), \alpha }} \le {\rm{t}} \le {\rm{\beta }}\)

Then using previous formula for the area under parametric curve, it is obtained that –

\({\rm{A = }}\int_{\rm{a}}^{\rm{b}} {{\rm{ydx}}} {\rm{ = }}\int_{\rm{\alpha }}^{\rm{\beta }} {{\rm{g(t)f'(t)dt}}} \)

Therefore, the result is obtained as \({\rm{A = }}\int_{\rm{\alpha }}^{\rm{\beta }} {{\rm{g(t)f'(t)dt}}} \).

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