/sci/ - Science & Math

[Anonymous No.5588523]

This question is really winding me up now please tell me what i am doing wrong /sci/!

My working:
y=sqrt(3x-5)
f ' (x)=1.5(3x-5)^-0.5
not sure what to do from here

[Anonymous No.5588531]

>>5588523
>please tell me what i am doing wrong /sci/!

You are posting high school homework.

[Anonymous No.5588539]

I've used primes for the x, y coords of the line.

y' - sqrt(3x-5) = 3/(2sqrt(3x-5))(x' - x)

sub in (x',y') = (0,0)

solve

[Suomynona !!rGxutZjWR7L No.5588545]

>tell me what i am doing wrong /sci/
You are trusting 4chan to help you pass a class

[Anonymous No.5588581]

>>5588539
expanded and simplified and got 2/3. obviously this isnt right unless i have to sub it into something else

[Anonymous No.5588591]

>>5588581
you expanded wrong

[Anonymous No.5588597]

>>5588523
differentiate the curve to get the equation of the tangent.
simultaneous solve the tangent and the curve

[Anonymous No.5588605]

>>5588523
Plug in x=0 and find the slope via the derivative, then plug in x=10/3 into the derivative, if they have the same slope, they both lay on the same tangent line. Bro, simple calculus

[Anonymous No.5588609]

>>5588581
that should be 10/3 retard

captcha: hetnin Harriet

[Anonymous No.5588629]

>>5588591
yeah i did i got 5/3 this time and i know what to do know thanks

[Anonymous No.5588634]

>>5588629
still wrong it is 10/3

[Anonymous No.5588637]

>>5588523
Your equation:
<span class="math">y(x) = \sqrt{3x - 5}[/spoiler]
Derivative:
<span class="math">y'(x) = \frac{3}{2\sqrt{3x - 5}}[/spoiler]
Tangent line:
<span class="math">T_x(s) = y(x) + y'(x)(s - x)[/spoiler]
You now impose that when s equals 0 the line equals 0(origin):
<span class="math">y(x) + y'(x)(0 - x) = 0 \Rightarrow y(x) -xy'(x) = 0 => y'(x) = \frac{y(x)}{x}[/spoiler]
Then you find the x that satisfies this equality:
<span class="math">\frac{3}{2\sqrt{3x - 5}} = \frac{\sqrt{3x-5}}{x} \Rightarrow 3x = 2(3x - 5) \Rightarrow 3x = 6x - 10 \Rightarrow 3x = 10 \Rightarrow x = 10/3[/spoiler]

[Anonymous No.5588676]

>be stuck on maths homework involving line integrals, integration about a curve, or whatever you call it
>check the rules to see if homework threads are allowed on /sci/
>Homework threads will be deleted and the poster banned
>decide to visit /sci/ to see what it is like anyway
>2 homework threads on front page, one an obvious troll, the other high school shit about gradients

nice moderation guys

[Anonymous No.5588732]

a straight line has an equation like y=k*x+d, d is 0 here because it goes through the origin. k is the gradient of the straight line. in that problem thats the same as the first derivative of your original equation.

then you just set both functions equal and simplify to get the answer.

y_tangent=k*x+d=y_original'*x+0

y_tangent=y_original

3x/(2sqrt(3x-5))=sqrt(3x-5)

x=10/3

why so complicated bitches?