Find d y / d x and d^2 y / d x^2 . For which values of t is the curve concave upward? x=t^3+1,

Find d y / d x and d^2 y / d x^2 . For which values of t is...

Key Concepts

Second Derivative for Parametric Curves

To find the second derivative d²y/dx² for a curve defined parametrically, one differentiates the first derivative dy/dx with respect to the parameter t and then divides the result by dx/dt. This process, expressed as d²y/dx² = (d/dt (dy/dx))/(dx/dt), provides insight into the curvature and concavity of the curve.

Parametric Equations

Parametric equations describe a curve by expressing both x and y as functions of a third variable, called the parameter (often denoted as t). This formulation allows for analysis of curves that may not be easily described by a single function y = f(x), and it is particularly useful when dealing with curves that have loops or vertical tangents.

Parametric Differentiation

In parametric differentiation, the derivative dy/dx is computed by taking the derivative of y with respect to the parameter t and dividing it by the derivative of x with respect to t. Formally, dy/dx = (dy/dt)/(dx/dt), provided that dx/dt is not zero. This technique leverages the chain rule to relate the rates of change between the variables.

Chain Rule

The chain rule is a fundamental principle in calculus used to differentiate composite functions. When applying the chain rule in the context of parametric equations, it connects the derivative with respect to one variable to the derivatives with respect to the parameter, allowing the computation of dy/dx from dy/dt and dx/dt.

Concavity and Second Derivative Analysis

The concavity of a curve is determined by the sign of its second derivative. If the second derivative is positive, the curve is concave upward on that interval; if negative, the curve is concave downward. This analysis is crucial in understanding the behavior of curves, including identifying points of inflection where the concavity changes.

Transcript

00:02 Alright, so this is what we are given, and we first want to look for the derivative of y with respect to x.

00:12 And that we know will be the derivative of y with respect to t all over the derivative of x with respect to t.

00:23 Well, the derivative of y with respect to t is the derivative of t squared minus t.

00:29 So that will be 2t minus 1 all over the derivative of the derivative.

00:37 Of x with respect to t, which is the derivative of t cubed minus 1, which is 3t squared, using the power rule.

00:59 So now we want to look for the second derivative, which we know be the derivative with respect to t of my first derivative, all over the derivative of x with respect to t.

01:32 Well, since i'm taking the derivative of my first one, i'm going to simplify it some.

01:41 By first noting, i can split this up into two fractions, combining exponents in the first one.

02:06 I get this minus, sorry, 1 3rd t to the negative 2.

02:25 So now i can easily take the derivative by simply applying power rule.

02:29 To this.

02:32 So that will be two -thirds times the derivative of t to the negative 1, which is negative t to the negative 2, minus 1 third times the derivative of t to the negative 2, which is negative 2, t to the negative 3, all over the derivative of x with respect to t.

02:56 Here's my x, and the derivative of that is 3t squared.

03:04 So now we should simplify this and make it look nicer.

03:10 I'm first going to do that by making my division into multiplication.

03:28 So i move my denominator to the top as so.

03:34 Then i can distribute it out.

03:38 One thing i want to notice right now is inside here i have a negative.

03:45 Sorry, i'm minusing a negative.

03:47 So those both become positive.

03:52 And now i'll take this and distribute it to the two terms in my sum to get two, i'm sorry, not two, six, two ninths because it's three times three.

04:16 Or negative two ninth times t to the negative four plus one ninth, negative or sorry.

04:44 I have a two right here.

04:45 So that's plus two -ninths times t to the negative five.

04:58 So i am going to pull out a two -ninth and a t to the negative five, which will leave me with negative t plus one.

05:16 In other words, this is two times one minus t all over nine, t to the fifth and this is my second derivative so now we'll test to see when this is concave up we know it will be concave up when my second derivative is positive now this can happen one of two ways so let's split it first both my numerator and my denominator can be positive so two times one minus t is greater than 0 and 9t to the 5th is greater than 0 or they can both be negative...

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