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Find the y-intercept of the tangent line to: -.4/ sqrt(3 + x) at

by Lo.Lee.Ta.
Tags: 4 or, line, sqrt3, tangent, yintercept
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Lo.Lee.Ta.
#1
Jan9-13, 06:12 PM
P: 209
1.Find the y-intercept of the tangent line to: y= -.4/√(3 + x) at [2.5, -.170560573084488]

2. So I thought the first step would be to find the slope of the tangent line.
I think we find the slope of the tangent line by taking the derivative.

So I am going to use the Quotient Rule to take the derivative.

[(√3 + x)(0) - (-.4)(.5(x + 3)^-1/2) * 1] / √(x + 3)^2

.2√(x + 3) / (x + 3) <----- This is the slope of the tan line to y= -.4/√(3 + x)

If we want to know the slope at x = 2.5, then wouldn't we plug in 2.5 for x?

So, [.2√(3 + 2.5)]/(2.5 + 3) = .085<----- This is the slope


Now, I thought we could substitute everything into y = mx + b in order to find out what b is.

-.1705606 = .085(2.5) + b

b = -.3831

3. But this is NOT the right answer!
Please tell me what I am doing wrong!
Thank you! :)
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Dick
#2
Jan9-13, 06:16 PM
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Quote Quote by Lo.Lee.Ta. View Post
1.Find the y-intercept of the tangent line to: y= -.4/√(3 + x) at [2.5, -.170560573084488]

2. So I thought the first step would be to find the slope of the tangent line.
I think we find the slope of the tangent line by taking the derivative.

So I am going to use the Quotient Rule to take the derivative.

[(√3 + x)(0) - (-.4)(.5(x + 3)^-1/2) * 1] / √(x + 3)^2

.2√(x + 3) / (x + 3) <----- This is the slope of the tan line to y= -.4/√(3 + x)

If we want to know the slope at x = 2.5, then wouldn't we plug in 2.5 for x?

So, [.2√(3 + 2.5)]/(2.5 + 3) = .085<----- This is the slope


Now, I thought we could substitute everything into y = mx + b in order to find out what b is.

-.1705606 = .085(2.5) + b

b = -.3831

3. But this is NOT the right answer!
Please tell me what I am doing wrong!
Thank you! :)
Your derivative is wrong. What is u^(-1/2)/u? You aren't combining the exponents correctly. u^(-1/2)=1/sqrt(u). Not sqrt(u).
SteamKing
#3
Jan9-13, 06:21 PM
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For calculating the derivative, instead of using the quotient rule, rewrite the equation:
y = -0.4/SQRT(x+3) as y = -0.4 * (x+3)^(-1/2)

Now, you have an eqn. of the form y = a u^n and you can use the power rule to get the derivative.

The Quotient Rule is useful when you are dealing with the ratio of two functions.

Mark44
#4
Jan9-13, 06:24 PM
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Find the y-intercept of the tangent line to: -.4/ sqrt(3 + x) at

Quote Quote by Lo.Lee.Ta. View Post
1.Find the y-intercept of the tangent line to: y= -.4/√(3 + x) at [2.5, -.170560573084488]

2. So I thought the first step would be to find the slope of the tangent line.
I think we find the slope of the tangent line by taking the derivative.

So I am going to use the Quotient Rule to take the derivative.
A better choice (because it's simpler and less prone to errors) would be write your equation as y = -.4 * (x + 3)-1/2
Quote Quote by Lo.Lee.Ta. View Post

[(√3 + x)(0) - (-.4)(.5(x + 3)^-1/2) * 1] / √(x + 3)^2
It's a good idea to keep track of what you're doing. For the above, you're working with the derivative, so this line should start with y' = or dy/dx = .

I see that you are including brackets around your numerator, so you've been listening. Good!

You have a mistake above that might just be a transcription error. At the beginning you have (√3 + x)(0) ...
That should be [(√(3 + x)(0)
Quote Quote by Lo.Lee.Ta. View Post

.2√(x + 3) / (x + 3) <----- This is the slope of the tan line to y= -.4/√(3 + x)
You have a mistake above. You should have gotten
dy/dx = .2(x + 3)-1/2/(x + 3)

Quote Quote by Lo.Lee.Ta. View Post

If we want to know the slope at x = 2.5, then wouldn't we plug in 2.5 for x?

So, [.2√(3 + 2.5)]/(2.5 + 3) = .085<----- This is the slope


Now, I thought we could substitute everything into y = mx + b in order to find out what b is.

-.1705606 = .085(2.5) + b

b = -.3831

3. But this is NOT the right answer!
Please tell me what I am doing wrong!
Thank you! :)
Mark44
#5
Jan9-13, 06:29 PM
Mentor
P: 21,397
To expand on what SteamKing said, you should NEVER use the quotient rule if either the numerator or denominator is a constant.

It's not incorrect to use it in this case, but it's more complicated, and the chances are greater of making a mistake.

For example, if f(x) = 2/(x2 + 1), write this as f(x) = 2(x2 + 1)-1 and use the chain rule to get f'(x) = 2(-1)(x2 + 1)-2(2x).
Lo.Lee.Ta.
#6
Jan9-13, 07:01 PM
P: 209
When I was working it out on paper, I left out the (-) on the -1/2 exponent!

Thanks for pointing that out!

So the Quotient Rule method is:

[.4(1/2(x + 3)^-1/2)]/(x + 3)

.2/(x + 3)^1/2 * (x + 3)

= .2/ (x + 3)^3/2

If I substitute the x = 2.5, then I get: .0155 as the slope.

I tried the Product Rule as well to see if I'd get the same thing.

So, -.4*-1/2(x + 3)^-3/2 + [(x + 3)^-1/2](0)

= .2(x +3)^-3/2

When substituting x = 2.5, I get the same slope of .0155


When I substitute everything into y = mx + b, I get: -.17056 = .0155(2.5) + b
b = -.209

...But this is not the right answer...
Do you know what could be wrong?
Is my derivative still wrong?
Thanks! :)
Dick
#7
Jan9-13, 07:54 PM
Sci Advisor
HW Helper
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P: 25,235
Quote Quote by Lo.Lee.Ta. View Post
When I was working it out on paper, I left out the (-) on the -1/2 exponent!

Thanks for pointing that out!

So the Quotient Rule method is:

[.4(1/2(x + 3)^-1/2)]/(x + 3)

.2/(x + 3)^1/2 * (x + 3)

= .2/ (x + 3)^3/2

If I substitute the x = 2.5, then I get: .0155 as the slope.

I tried the Product Rule as well to see if I'd get the same thing.

So, -.4*-1/2(x + 3)^-3/2 + [(x + 3)^-1/2](0)

= .2(x +3)^-3/2

When substituting x = 2.5, I get the same slope of .0155


When I substitute everything into y = mx + b, I get: -.17056 = .0155(2.5) + b
b = -.209

...But this is not the right answer...
Do you know what could be wrong?
Is my derivative still wrong?
Thanks! :)
Now I don't see anything wrong with what you did. It looks correct to me. I get b=(-0.209324).
Lo.Lee.Ta.
#8
Jan10-13, 11:43 AM
P: 209
@Dick - Thank you so much for always helping me! :) You are kind.

I used the extra digits you found, and NOW the computer counts it as correct!
Thanks! :D


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