Equations of Tangent Lines: Trig-1

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For the parametric equations trig parametric equations

,
find the equation of the tangent line at

How do we proceed?

We can use the point-slope form of the equation of the tangent line because we can find the coordinates of the point from the value of t and can find the slope at that point.

As a review, state the point-slope form for a line of slope m and passing through the point (x_o, y_o).

How do we find the slope?

From

, which we can obtain from the parametric equations.

State the general relationship among

general equation of the derivative from parametric equations

Determine

Substitute these results in

derivative of the parametric trig equations

Evaluate this at

We get

Now we need the value of the coordinates that correspond to this value of t. Determine the x-coordinate.

Determine the y-coordinate.

Use this data in the point-slope equation to find the equation of the tangent line at t = 0.

At t = 0,

Simplify this equation.

The equation of the tangent line at t = 0 is equation of the tangent line to the graph of the parametric trig equations

To check these results, plot the parametric equations for the interval 0 < t < 1.57. Then check your graph by clicking “Next”.

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Add the point for . Then check your graph by clicking “Next”.

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Add the tangent line at that point. Then check your graph by clicking “Next”.

alt="Graph of the parametric trig equations showing the tangent. Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag!

Does the tangent line go through the appropriate point on the graph?

Yes: through

Is the slope correct?

Yes, it is negative and gradual (-1/2).

Is the y-intercept correct?

Yes, it is at y = 1.

The end. If you found this helpful and would recommend that I create more pages like this one, please let me know: Email to John Taylor

General Contents

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