Find an equation for the line tangent to the curve at the point defined by the given value of t

step 1 Okay, I want to find DY by DX when T is negative one -sixth. So first step, let's find DX by DT,

Find an equation for the line tangent to the curve at the...

Key Concepts

Parametric Equations

Parametric equations define a curve by expressing both x and y as functions of a third variable, usually denoted by t. This representation allows for describing curves that are difficult to express using standard Cartesian equations, and it is especially useful when analyzing motion or tracing closed curves.

Parametric Differentiation

Parametric differentiation involves computing derivatives of curves defined by parameters. Instead of finding dy/dx directly, one calculates dx/dt and dy/dt and then finds dy/dx as the ratio (dy/dt)/(dx/dt). This method is essential for understanding the slope and behavior of parametric curves at given points.

Tangent Line

The concept of a tangent line pertains to a straight line that touches a curve at a single point and has the same instantaneous direction as the curve at that point. To determine the tangent line for a parametric curve, one calculates the derivative dy/dx at the specific value of t, then uses the point-slope form of the line equation.

Second Derivative for Parametric Equations

The second derivative d²y/dx² of a parametrically defined curve provides information about the curvature or concavity of the curve. This second derivative is found by differentiating the first derivative dy/dx with respect to t and then dividing by dx/dt, applying the formula d²y/dx² = (d/dt(dy/dx))/(dx/dt).

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. When dealing with parametric equations, the chain rule is applied to differentiate one dependent variable with respect to another by relating their derivatives through the parameter t, especially when computing second derivatives.

Transcript

00:01 Okay, i want to find dy by dx when t is negative one -sixth.

00:05 So first step, let's find dx by dt, differentiate sine 2 pi -t, i get cosine 2 -t, i get cosine 2 -t times 2 -py by the chamberl.

00:21 And then if i work out d -y by d t, i get negative sine 2 -pt, cosine, cosine, negative sign and then again the chain rule negative 2 pi so now i want to work out d y by d x and that will be d y by d t times d t by d x again the chain rule so that will equal negative two pi sine two pi t times one over two pi cosine 2 pi t and then the 2 pi is cancel out and we're left with sign of a cause is tan so minus tan 2 pi t now when t is negative 1 6th d y by d x will be negative tan and negative is it negative yes negative 2 pi over 6 replace the t by negative by negative 1 6th and that becomes, well, a tan of minus theta is minus tan theta.

02:15 So overall, this comes down to tan of pi by 3.

02:24 And tan pi by 3 is root 3.

02:29 Now, my tangent is y equals mx plus b, and the m is root 3, x, and the m is root 3, x, and plus the b and let's find out a point on this tangent is the point where it meets the curve so back to here then the x and y values will be this so this will be sign of by plug in negative one -six i will get negative pi by three and that will equal minus sine 60 so minus root 3 over 2 and then we have the cosine of negative pi by 3.

03:20 Same thing here as cosine plus pi by 3.

03:25 So just one half.

03:29 So we now know that negative root 3 over 2 and 1 half is a point on this tangent.

03:52 1 half equals root 3, negative root 3 over 2 plus b.

03:59 So 1 half equals 3 over 2, minus 3 over 2 plus b.

04:09 So b then will be 2.

04:13 So my answer is y equals root 3 of x plus 2...