Question

A straight line perpendicular to and passing through the point of tangency of the tangent line is called the normal to the curve. Find an equation of the tangent line and the normal to the curve y = x3 - 3x + 3 at the point (2, 5). tangent line y = normal line y =

          A straight line perpendicular to and passing through the point of tangency of the tangent line is called the normal to the curve. Find an equation of the tangent line and the normal to the curve y = x3 - 3x + 3 at the point (2, 5).
tangent line     y =   
normal line     y =

Show more…

Added by Gregory B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Ekaveera Kumar

03/05/2021

Step 1

The derivative of y with respect to x is given by y' = 3x^2 - 3. Show more…

Show all steps

lock

Thanks for your feedback!

A straight line perpendicular to and passing through the point of tangency of the tangent line is called the normal to the curve. Find an equation of the tangent line and the normal to the curve y = x3 - 3x + 3 at the point (2, 5). tangent line y = normal line y =

Play audio

Feedback

Powered by NumerAI

Ekaveera Kumar and 97 other subject Calculus 1 / AB educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 The given curve is y is equal to x a cube minus 3x plus 5.

00:10 We need to find the equation of tangent at 3 .23 and also we need to find the equation of normal.

00:19 Geometrically, the slope of a tangent, slope of the tangent at a point p is derivative of the curve at that point p.

00:36 So let's find the derivative of this curve.

00:39 So what is d .y by dx? derivative of x cube is 3x squared minus derivative of 3x is 3.

00:46 Now, what is the derivative of this curve at this point 3 comma 23? basically, x value is 3, right? because x coordinate is 3 here.

00:55 X, 0 is 3.

00:57 So let's substitute x0 value here.

01:01 So 3 into x0 is 3 square minus 3.

01:04 So how much? 27 minus 3.

01:06 So 24 is the slope of the tangent.

01:09 So this is the slope of the tangent.

01:12 So tangent is also passing through 3 .23.

01:15 So slope is 24 and the line passes through, the tangent passes through 3 .23.

01:28 So what will be the equation of straight line when you know slope and point, y minus y 1 is equal to m into x minus x1.

01:37 So, the equation of tangent is y minus 23 is equal to 24 into x minus 3.

01:43 So when you simplify this, y minus 23 is equal to 24x minus 24 into 3, 72.

01:50 So you can put it in proper form, 24x minus y and 23 minus 72...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later