Find the equation of the tangent line at t=(π)/(4) to the cycloid generated by the unit circle w

Find the equation of the tangent line at t=(π)/(4) to the...

Key Concepts

Parametric Equations

Parametric equations represent curves by expressing the coordinates of the points on the curve as functions of a parameter, usually denoted by t. This approach allows for a more flexible description of curves that might not be easily described by a single function of x or y, and it facilitates the analysis of motion and other dynamic processes along the curve.

Cycloid

A cycloid is the curve traced by a fixed point on the circumference of a circle as it rolls along a straight line without slipping. This classic example of a parametric curve exhibits interesting geometric properties and has applications in various fields like physics and engineering, as well as in the study of brachistochrone problems.

Differentiation of Parametric Equations

Differentiation of parametric equations involves computing the derivatives of the coordinate functions with respect to the parameter and using these to determine the derivative of y with respect to x. This is typically accomplished by calculating dy/dt and dx/dt, and then finding the slope of the tangent to the curve by taking the ratio (dy/dt)/(dx/dt), assuming dx/dt is nonzero.

Tangent Line to a Parametric Curve

The tangent line to a curve at a given point is a straight line that just touches the curve at that point and has the same instantaneous direction as the curve. For curves defined parametrically, once the point on the curve and the slope (determined by the derivatives of the parametric functions) are known, the point-slope form of a line can be used to write the equation of the tangent line.

Transcript

00:01 All right, here's the equations for the cycloid, and it wants me to find the equation of the tangent line that's tangent to the cycloid at t equals pi over four.

00:12 Okay, to find the equation of a line.

00:15 I need two things.

00:16 I need a point, and i'm going to get that with the t equals pi over four, and i need a slope, which i'm going to get by taking the derivative, or finding the derivative.

00:29 All right, so we'll work on those two things, and then i'll write that.

00:32 The equation.

00:33 The point is named xy and it's at t equals pi over four so it is pi over four minus the sign of pi over four because it was t minus the sign right okay and then the y value is one minus the cosine of pi over four pi over four is 45 degrees so both the sign and cosine are one over the square root of two or square root of two over two okay, and just to make them nice, i'm going to get a common denominator.

01:12 So i get pi minus two square root of two over four, and two minus the square root of two over two for this one.

01:25 Okay, slope.

01:29 To find the slope, i need to find d, y, dx.

01:32 And remember, its formula is dy, d, whatever variable we're using for the parameter, and dx d.

01:40 T so the derivative of one minus the cosine of t is derivative of the cosine is minus sine so it's sine t over then the derivative of x will be one minus cosine t okay now that's not the slope of the tangent yet that's just the derivative to make it into the slope of the tangent you have to plug in something it has to be a number so here what we're plugging in is t equals pi over four.

02:16 So again it's the square root of two over two and one minus the square of two over two and then let's multiply everything by two to get rid of those fractions.

02:30 So we get the square root of two over two minus the square root of two.

02:39 Okay, i'm gonna use it like that and then i will fix it up if i have to.

02:46 Oh yeah, that's gonna be bad.

02:48 Let's go ahead and fix it up square to 2 over 2 minus the square root of 2 and the fixing up is getting that 2 minus the square to 2 out of the bottom and to do that i'm going to multiply by the conjugate 2 plus the square root of 2 so on the top i get two square roots of 2 plus 2 and on the bottom i get 4 minus 2 because they're conjugate you just square the first one square the second one put a minus sign in between.

03:24 So get two squareds of two plus two over two.

03:30 Factor the two out of the top, square of two plus one over two.

03:36 All right, so m tangent is the square root of two plus one...

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