![]() Thinking of t as time, then both the velocity vector and accelerations vectors are simply derivatives in each component.įor more information about vector functions, check out this AP Calculus BC Review: Vector-Valued Functions Example - Velocityįind the velocity vector at t = 1 for an object traveling according to the parametric function x = t 2 – 2 t + 1, y = – t 2 + 2.įirst find the derivative of each component function. ![]() The idea is to think of a parametric function as a vector. There is another interpretation of the derivative that allows you to compute the velocity of an object traveling along a parametric curve. 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.
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