Question

at what values of x do the graphs of y= x^2 and y=-sqrt(x) have perpendicular tangent lines?

Answer 1

have perpendicular tangent lines to each other? (saying that the tangents are perpendicular?)

Answer 2

lets start from finding the derivative of the entire function.

Answer 3

I mean of the 1st function

Answer 4

are you supposed to graph, or just find it on some other way?

Answer 5

other way, jdoe

Answer 6

hmm

Answer 7

y'=2x

Answer 8

We will need the derivative of both of the functions:
lemme do a notation rearrangement:

\(\normalsize\color{royalblue}{ \rm f(x)= x^2 }\)
\(\normalsize\color{royalblue}{ \rm g(x)=-\sqrt{x} }\)

Answer 9

and the second function y'=-1/2(x)^-1/2

Answer 10

yes, \(\normalsize\color{royalblue}{ \rm f'(x)= 2x }\)

Answer 11

and \(\normalsize\color{royalblue}{ \rm g'(x)= \frac{\Large 1 }{\Large2\sqrt{x}} }\)

Answer 12

Note that:

\(\normalsize\color{royalblue}{ \rm f'(x)=2x}\) and \(\normalsize\color{royalblue}{ \rm g'(x)= \frac{\Large 1 }{\Large2\sqrt{x}} }\)

are the slopes. of both of the functions:

Answer 13

you need to find a value of a, (for x), so that

\(\normalsize\color{royalblue}{ \rm g'(a)=~- \frac{\Large 1 }{\Large f'(a)} }\)

or,

\(\normalsize\color{royalblue}{ \rm f'(a)=~- \frac{\Large 1 }{\Large g'(a)} }\)

it is all same

Answer 14

\(\normalsize\color{royalblue}{ \rm f'(x)=~- \frac{\Large 1 }{\Large g'(x)} }\)


\(\normalsize\color{royalblue}{ \rm 2x=~- \frac{\Large 1 }{\Large \frac{\Large 1 }{\Large 2\sqrt{x}}} }\)

and on...

Answer 15

oh my bad.

Answer 16

you function g(x) is negative so the derivative is negative.

\(\normalsize\color{royalblue}{ \rm 2x= \frac{\Large 1 }{\Large \frac{\Large 1 }{\Large 2\sqrt{x}}} }\)

Answer 17

I am setting the perpendicular slope equation, substituting the derivative of the functions into that formula. (2 minus on the left side make it positive)

Answer 18

ok

Answer 19

got lost after you function g(x) is negative so the derivative is negative....

Answer 20

So I look for the derivatives, then set one of them equal to the perpendicular slope of the other one?

Answer 21

You know that: m=-1/m are prep slopes ? (excuse me for these words, mathematicians)

Answer 22

2x= -2sqrt(x)/1

Answer 23

close

Answer 24

you derivative is negative, and you got another negative in the m=-1/m

Answer 25

your deriv.* ...

Answer 26

I mean the derivative for -sqrt{x}

Answer 27

\(\normalsize\color{royalblue}{ \rm f'(x)=-\frac{\Large 1}{\Large g'(x)} }\)

Answer 28

perpendicular slopes, correct?

Answer 29

yes

Answer 30

\(\normalsize\color{royalblue}{ \rm 2x=-\frac{\Large 1}{\Large \frac{\LARGE 1}{\LARGE -2\sqrt{x}}} }\) plugging the derivatives in...

Answer 31

you got it right that time, for everything besides the sign... see why?

Answer 32

wait how is 2sqrt(x) negative?

Answer 33

you had negative sqrt{x} in the begginging, no?

Answer 34

ohh yes

Answer 35

yes, see why you get:

\(\normalsize\color{royalblue}{ \rm 2x=-\frac{\Large 1}{\Large \frac{\LARGE 1}{\LARGE -2\sqrt{x}}} }\) >>> \(\normalsize\color{royalblue}{ \rm 2x=2\sqrt{x}}\)

Answer 36

and then the x, is?

Answer 37

sqrt x

Answer 38

?

Answer 39

what do you mean?

Answer 40

\[\sqrt{x}\]

Answer 41

I see, but what do you want to tell me?

Answer 42

All I did so far is that I set the derivatives of f(x) and g(x):


(
where: ~ \(\normalsize\color{royalblue}{ \rm f(x)=x^2}\)
~ \(\normalsize\color{royalblue}{ \rm f'(x)=2x}\)

~ \(\normalsize\color{royalblue}{ \rm g(x)=-\sqrt{x}}\)
~ \(\normalsize\color{royalblue}{ \rm g'(x)=-1/(~2\sqrt{x}~)}\)

Answer 43

set the derivatives (the slopes of f(x) and g(x) into a form of:

m = -1/m

Answer 44

since I need to find the x, where the slopes of the functions are perpendicular.
(and the slopes are the derivatives)

Answer 45

ok -1/1/-2sqrt(x) and -1/2x

Answer 46

no, you need the slopes of the f(x) and g(x) perp to each other:

that is why I set: `f'(x) = - 1 / g'(x)`

makes sense?

Answer 47

lost, or with me?>

Answer 48

ok so \[2x=\frac{ -1 }{ \frac{ 1 }{ -2\sqrt{x} } }\]

Answer 49

@SolomonZelman

Answer 50

yes

Answer 51

and then solve for x.

Answer 52

this x value will be where the slopes of the 2 functions are perpendicular to each other.

Answer 53

(note: exclude extraneous solutions)

Answer 54

lost me, or can solve for x?

Answer 55

I get x= sqrt(x)

Answer 56

yes, that means that x=?

Answer 57

there are 2 solutions, what are they?

Answer 58

\[\pm \sqrt{x}\]

Answer 59

no plus minus

Answer 60

and secondly,
x=\sqrt{x}

doesn't that imply that x=1 or 0?

Answer 61

0 won't work, because it is not in the g'(x), so we are left with 1.

Answer 62

so the tangents will be perpendicular at x=1.

Answer 63

want to find these tangent lines and graph them to see if it is true?

Answer 64

yes :)

Answer 65

We have:

\(\normalsize\color{royalblue}{ f(x)=x^2 }\)
\(\normalsize\color{royalblue}{ f'(x)=2x }\)

\(\normalsize\color{teal}{ g(x)=-\sqrt{x} }\)
\(\normalsize\color{teal}{ g'(x)=-1/(~2\sqrt{x}~) }\)

Answer 66

find the tangent line to f(x), at x=1. (can you do that? )

Answer 67

steps for this:
Find \(\normalsize\color{royalblue}{ f'(1) }\)
Find \(\normalsize\color{royalblue}{ f(1) }\) (y coordinate of the point)
Use the point slope form.

Answer 68

y-1=2(x-1)
y=2x-1

Answer 69

yes, we can leave in a point slope form, ain't important

Answer 70

=D

Answer 71

can you find the slope of the tangent line, to the curve g(x), at x=1 ?

Answer 72

y+1=-1/2x-1/2

Answer 73

without expanding, so that I we can clearly see the point plugged: y+1=-1/2(x-1)

Answer 74

since g(1) is the y coordinate of the point, and 1 is the x coordinate:
g(1)=-sqrt(1) = -1

the point is (1,-1) . and with a slope -1/2, that gives as: the above line.

slope of -1/2 is correct;) f'(1)= -1/(2sqrt(1)) = -1/2

Answer 75

So tangents at \(\normalsize\color{royalblue}{ x=1 }\)

For \(\normalsize\color{royalblue}{ f(x) }\): \(\normalsize\color{royalblue}{ y-1=2(x-1) }\)

For \(\normalsize\color{teal}{ f(x) }\): \(\normalsize\color{teal}{ y+1=\frac{1}{2}(x-1) }\)

Answer 76

There is the graph of it:

https://www.desmos.com/calculator/efm9gxyiam