Therefore, using the point-slope form, the equation of the tangent line is: Now we also need to know what the x– and y-coordinates are for the point in question. What is the equation of the tangent line at t = π/6 for the parametric function x = 3 cos t, y = 3 sin t?ĭy/ dx = (3 cos t)/(-3 sin t) = -cos t / sin t. In fact, you’ll have to take the derivative of both. Should you take the derivative of f( t) or g( t)? We must be careful, because there are two equations to deal with. But how do you find the derivative of a set parametric equations? As you know, the derivative measures slope. Now that you have seen some graphing, let’s talk about slope. Next, plot these points on a coordinate plane.įinally, connect the dots in order of increasing t. Remember, use both positive and negative values to get a good sense for how the function behaves. Graph the parametric function defined by x = t 2 – 2 t + 1 and y = – t 2 + 2.īecause there was no range specified for t, let’s just pick a few easy numbers to work with. Then connect the dots in the order of increasing t. For each sample t, plug it into x = f( t) and into y = g( t) to find out the corresponding x– and y-coordinates.Otherwise, just pick a wide range of values, both positive and negative. If there is a given range of values, a ≤ t ≤ b, then you must stick to values within that interval. There is a straightforward way to graph any parametric function. This is about the best I could ever manage on an Etch-A-Sketch.
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