Use the method in Example 3 to find (a) the slope of the curve at the given point P, and (b) an

Use the method in Example 3 to find (a) the slope of the...

Key Concepts

Differentiation

Differentiation is the process in calculus that deals with finding the derivative of a function, which measures the function's instantaneous rate of change with respect to its independent variable. This concept is central in understanding how a function behaves locally near a point, making it a fundamental tool in solving problems related to slopes and rates of change.

Tangent Line

A tangent line to a curve at a given point is the line that best approximates the curve near that point. It has the same slope as the curve at the point of tangency, meaning it reflects the instantaneous rate of change of the function at that point. Understanding tangent lines helps in approximating functions and analyzing their behavior locally.

Point-Slope Form

The point-slope form is a method used to express the equation of a line when the slope and a point on the line are known. This form, given by y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, is particularly useful in the context of finding the equation of a tangent line to a curve at a specific point.

Transcript

00:02 Hello everyone.

00:04 Now let us look at an example problem to find the slope of a given curve at a given point p and also find the equation of a tangent line at p.

00:17 So now let us look at the given example and now let us note down the information that is given.

00:26 We are given with the equation of a curve y equals x square minus 3 and we're also given with the point p 2 comma 1.

00:39 Now let us note down what is to be found.

00:47 So we need to find the slope of the given curve at p that is this point and now we also need to find the equation of the tangent that passes through the point b.

01:34 Now let us find answers to these questions.

01:41 Now let us look at how to solve or how to find the slope of the given curve at p.

01:49 So first things first we can just we have to see if the given point p 2 comma 1 does it actually lie on the given curve y equals x square minus 3.

02:07 So we can just substitute, we can just substitute y equals 1 and x equals 2 in equation 1.

02:26 And we find that 1 is equal to 2 square minus 3 which is equal to 1 and hence the equation 1 is satisfied.

02:36 So we can say that point p lies on the given curve.

02:50 So now we need to find the slope of the given curve and how do we find that? the slope of the given curve is found using the instantaneous slope which is found using the first derivative and evaluated at the given point.

03:20 So to start off with that we need to first familiarize ourselves with a few standard...

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