Assuming that the equations in Exercises 63-66 define x and y implicitly as differentiable funct

step 1 Welcome to this lesson. In this lesson we'll be finding the slope 4, the implicit function S and

Assuming that the equations in Exercises 63-66 define x and...

Key Concepts

Implicit Differentiation

Implicit differentiation is used when functions are defined by equations in which the dependent variable is not isolated on one side. Instead of expressing the variable explicitly, you differentiate both sides of the equation with respect to the independent variable, using appropriate differentiation rules, and then solve for the derivative. This method is crucial for dealing with equations that define variables implicitly.

Chain Rule

The chain rule is a fundamental differentiation technique used to differentiate composite functions. When differentiating a function that involves another function of the independent variable, you take the derivative of the outer function evaluated at the inner function and multiply it by the derivative of the inner function. This is essential when differentiating expressions like sqrt(5 - sqrt(t)) or similar nested functions.

Parametric Equations

Parametric equations describe a curve by defining both x and y as functions of a common parameter, often noted as t. This representation is particularly useful for curves that are difficult to express as a single explicit function. Analyzing such curves involves computing derivatives with respect to the parameter, which then allows you to determine properties such as the slope of the tangent.

Differentiation with Respect to a Parameter

When functions are given in terms of a parameter, differentiating with respect to that parameter involves computing the derivatives dx/dt and dy/dt. These derivatives quantify how x and y change as the parameter varies. This approach is important for finding the slope of a curve defined parametrically by linking the rate of change of y to the rate of change of x.

Slope of a Parametric Curve

The slope of a curve defined by parametric equations is determined by the ratio of the derivative of y with respect to the parameter t to the derivative of x with respect to t, expressed as dy/dx = (dy/dt)/(dx/dt). Evaluating this ratio at a specific parameter value gives the slope of the tangent to the curve at that point, which is a key aspect in understanding the curve's behavior.

Transcript

00:01 Welcome to this lesson.

00:02 In this lesson we'll be finding the slope 4, the implicit function s and y, at a point where t is equal to 4.

00:14 So the first thing we'll do is that because s depends on t, we'll find the differential of s with respect to t.

00:26 So i'll write the x in another form so that becomes very easy to do with the difference.

00:37 It the radical square root i'm replacing it but half so the s the t is equal to the half comes out then a 5 t to the bar half then because i brought the half forward i'm subtracting one from it making minus then after that i'm differentiating what is in a bracket so when i differentiate 5 minus square root of t i would get 1 over 2 square root of t okay so bringing all of them forward i'll have 1 over 4 that is a half here and i have there making 1 over 4 then 1 all over now i have square root of t then square root of five minus all the halves are square roots okay so instead of saying square root of t i can write t to the power half okay and if the half is negative it means that it comes down to the denominator side all right so the whole thing is better written as negative one, four.

02:33 So d s d t at t is equal to four.

02:39 When i put the value of t as four, this is what i have.

03:06 So two times four, then five minus two.

03:15 So that gives eight negative one over eight square root of so that is the x the t when t is equal to four.

03:36 In the same way, let's deal with the differential of why with respect to t.

03:48 Ok.

03:51 So y is equal to square root of t all over t minus 1.

03:59 When we write it in a very nice way so that we can deal with them.

04:06 You see the nature of the question and use the appropriate differentiation method...