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

Find an equation for the line tangent to the curve at the point defined by the given value of $t$. Also, find the value of $d^{2} y / d x^{2}$ at this point.
$$x=\sec t, \quad y=\tan t, \quad t=\pi / 6$$

Key Concepts

Parametric Equations

Parametric equations define a curve by expressing the coordinates of the points on the curve as functions of a parameter. This approach allows for a detailed study of the trajectory and properties of the curve by analyzing how x and y change with respect to that parameter.

Differentiation of Parametric Functions

When a curve is defined parametrically, the derivative of y with respect to x is obtained using the chain rule. Specifically, dy/dx is computed as (dy/dt) divided by (dx/dt). This method is essential for understanding the rate of change and slope of the curve at a given point.

Second Derivative in Parametric Form

The second derivative d²y/dx² measures the curvature of the curve. It is calculated by differentiating dy/dx with respect to the parameter t and then dividing by dx/dt, which provides insight into the concavity and the nature of the curve’s bending at that point.

Tangent Line to a Curve

The tangent line at a given point on the curve represents the line that best approximates the curve near that point. Its equation is derived using the point-slope form, where the slope is given by dy/dx evaluated at the specific parameter value, and the point corresponds to the coordinates obtained from the parametric equations.

Transcript

00:01 Okay, here i want to find a tangent to the curve given by these two equations here at the point where t is pi by 6.

00:13 So first step, find dy by dx.

00:17 To do that, let's work out dx by dt first.

00:22 D differentiates sec.

00:24 I get sec tan, so sec t tan t.

00:29 Then work out dy by dt.

00:32 Tan becomes sec square t and then to work out d y by d x that's by the chain rule d y by d t times d t by d x which is sec square t and then one over sec t tan t so the sec cut is out and we're left with sec t over tan t and sec is 1 over cos tan is sign over cause so the end result then is 1 over sign t or cosec t if you like now so when t then is pi by 6 d y by d x will be 1 over sine pi by 6 or sign pi by 6 is sign 30 is one half, so 2 is the slope.

02:01 So we now have then y equals mx plus b is my tangent equation and we know m is 2.

02:11 To find b, i want a point on the line.

02:15 That's the point of contact with the curve and that occurs where t is pi by 6.

02:22 So sec pi by 6 and tan.

02:29 Pi by 6 is the point i need.

02:33 And sec pi by 6 is 1 over cos pi by 6.

02:39 So 1 over root 3 over 2 or 2 over root 3.

02:49 Tan pi by 6, tan 30 is 1 over root 3.

02:55 So the x value is 2 over root 3 and the y value 1 over root 3.

03:01 So this point then lies on the tangent and the curve.

03:11 So we have then that 1 over root 3 equals twice 2 over root 3 plus b.

03:22 So b therefore is going to be negative 3 over root 3, which i can write a bit simpler as negative root 3 over 3.