Find d y / d x and d^2 y / d x^2, and find the slope and concavity (if possible) at the given va

Find d y / d x and d^2 y / d x^2, and find the slope and...

Key Concepts

Second Derivative in Parametric Form

The second derivative, d²y/dx², provides information about the concavity of the curve and is computed by differentiating dy/dx with respect to the parameter and then dividing by dx/d?. This process, expressed as d²y/dx² = (d/d?(dy/dx)) / (dx/d?), requires careful application of the chain rule to capture the curvature behavior of the curve.

Trigonometric Differentiation

When the functions in the parametric equations involve trigonometric terms, knowing the derivatives of these trigonometric functions (such as the derivative of sec? and tan?) is crucial. Trigonometric differentiation permits the accurate computation of the slopes and curvatures for curves defined using trigonometric functions, which often appear in real-world applications.

Chain Rule

The chain rule is a fundamental tool in calculus that allows the differentiation of composite functions. In the context of parametric curves, it is used to differentiate the first derivative (which is itself a function of the parameter) with respect to x, by relating the rate of change with respect to the parameter to the rate of change with respect to x.

Parametric Equations

Parametric equations represent curves by giving both x and y as functions of a third variable, commonly referred to as the parameter. This representation is particularly useful for describing curves that are difficult to represent with a single function, as it allows more flexibility in defining the behavior and properties of the curve.

First Derivative in Parametric Form

The first derivative of a curve expressed with parametric equations is found by dividing the derivative of y with respect to the parameter by the derivative of x with respect to the parameter, i.e., dy/dx = (dy/d?) / (dx/d?). This ratio gives the slope of the tangent line to the curve at any point, indicating the instantaneous rate of change of y with respect to x.

Transcript

00:01 Here we are given the parameter equations x is equal to 2 plus sec theta y is equal to 1 plus 2 tan theta and we're given the theta value pi over 6 and we have to compute the slope and determine the concave at the given theta value now now first we have refined dx the theta which will be the constant goes the derivative of sec is sec theta ton theta and here divide theta the one also goes and we have the derivative of two tan theta is two sec squared theta and our dy dyx is given by d -theta divided by d -x d -theta and we have two sec squared theta over sec theta turn theta and the sec cancel out leaving two sec theta over turn theta but we know that sec theta is one over cos theta apply by turn theta and here we are dividing right with turn theta so it's the same as multiplying with one over tan theta which will be course over sine theta like this so these two cancel out and we have our d -y -d -x being equal to 2 over sine theta which is 2 -cosec theta and this is our d -y -d -x our second derivative d squared y the x squared here it would be better actually to to have this as let's put it here we have our 2x as 2 over over sine theta therefore using the quotient rule the 2 is u so t u t theta is equal to 0 t v t theta is equal to cost theta therefore and we know that the second derivative will be t d theta of t y d x divided by d x d theta now first we find t d theta of t d theta using the question rule where v is sine theta applied by d u is 0 minus u 2, which 2, apply by dv, which is cos theta, over v squared, is sine squared theta.

03:46 And here we get minus 2, cost theta over sine squared theta.

03:56 And to finish and find d squared y, d x squared, we then divide by d.

04:04 X t theta so we have the squared y the x squared is equal to minus two cost theta over sine squared theta divided by d x t theta which is sec theta tan theta and this is the same as minus two cost so cost theta over sine square theta and say multiply by one over turn theta one over sec theta turn theta and that gives us so if we say one over sec theta 10 theta.

05:02 Now that sec is 1 over cos...

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