Problem

Use implicit differentiation to find an equation …

01:46

Question

Answered step-by-step

Problem 30 Easy Difficulty

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.
$ x^{\frac {2}{3}} + y^{\frac {2}{3}} = (-3 \sqrt{3}, 1), $ (astroid)

Video Answer

Solved by verified expert

Textbook Answer

Official textbook answer

Video by Doruk Isik

Numerade Educator

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Watch More Solved Questions in Chapter 3

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Video Transcript

In this problem, we are asked to use implicit differentiation to find the equation of tangent line to the given curve at just given point negative 3 state, 3 and 1. We know that equation of tangent line will be able to form y minus 1 is equal to the derivative of this function. Alice at given point multiply by x, minus x 9. So, in order to find the equation that we first need to find derivative lent at this given point all right: let's take derivative of this function with respects x. First, we have 2 third x, negative 1, third plus 2 or 3 y negative 1 third times y prime is equal to 0. So from this we see that y prime is the negative x over y to the power negative 1. Third point: since we know what x notariat is, we can calculate derivative at this given point so then that will be the derivative it will at that, given point will then be negative of x is 3 square 3, so that is negative 3 square root of 3 Times negative to 1. Third, that is equal to square root of 3, cube to the power negative 1 third, and that is equal to 1 over 33. Then, since now we know derivative at a given point, we can find the question of this line. That will be y. Minus 1 is equal to 1 over square root of 3 x, plus 3 square root of 3, and from this we dont fine equation to be square root of x. Sorry, we find the equation to be 1 over square root of 3 x. 1, over square root of b x, plus 4 pi.

Get More Help with this Textbook

James Stewart

Calculus: Early Transcendentals

View More Answers From This Book

Find Another Textbook

Study Groups

Study with other students and unlock Numerade solutions for free.

Problem