Find equation of tangent line and normal line?
Value of y' at 0.01 is = 1/25, I think.
@Miracrown
\[y-y_1=M(x-x_1)\]Where M is the slope of the tangent line
Did you get the function for y' ?
Yes, well, I got the derivative of the first y equation thing.
what di dyou get?
\[y'(x)=\frac{ \frac{ x+9 }{ 2\sqrt{x} }-\sqrt{x} }{ (x+9)^2 }\]
\[y ' = \frac{ -(x-9) }{ 2\sqrt{x}*(x+9)^2 }\]
equivalent, yeah? maybe yours is easier to work with xD
i just put it in my calculator, lol sorry
quotient rule, bottom times derivative top minus top times derivative bottom, all over bottom squared
Why do we need to do the quotient rule if we already have the derivative?
ooh right, I see what you're saying. Yep, that's what I did haha
ok, the point given on the function is (x,y) = (1 , 0.1) use that in the derivative, and y' will tell you the slope of the tangent line there at the point
like plug x into the derivative?
yeah figure out f ' (x) =
k, well I think f(1) = 1/25
yes, the value of the derivative at any point is the slope of a tangent line to the function at tha tpoint, so the tangent line has slope 1/25, and goes through the given point (x,y) make a line equation from that
the normal is perpendicular to a tangent at some point, just use the perpendicular slope to 1/25 and the same point for the normal line
hmm is it y = 1/25x-0.14?
tangent line : y - y1 = f '(x) *(x - x1) normal line, same, cept slope is -1/f'(x)
or y= 0.04x-0.14 for the tangent line
so the normal would be.. y=-25x-0.14? o.o
just miscalculated i think ... y - 0.1 = f '(x) *(x - 1)
I think your sign is off y - 0.1 = (1/25)(x-0.1) y - 0.1 = x/25 - 1/250
y = x/25 - 1/250 + 1/10 = 0.04x + 0.096
you can always graph everything and check if they look right, https://www.desmos.com/calculator
Where did you get that big 250?!
gah im wrong
x,y) = ( 1 , 0.1)
tangent - x/25 - 0.06
y - 0.1 = (1/25)(x-0.1)**** y - 0.1 = x/25 - 0.04
OMG
hah
y - 0.1 = (1/25)(x-1)**** y - 0.1 = x/25 - 0.04
when you add 0.1 to 0.04 it equals 0.14 though..
y = x/25 - 0.04 + 0.1
so youre actually subtracting .04 from .1
ooh I see i see.
And then the normal is the same, except -1/(1/25)
Which comes out to..
-25x-0.06 ??
lol yeah -1/25 - 1/10 = 3/50 or about 0.06
plus 1/10 omg
lol long day for everyone
lol yeah -1/25 + 1/10 = 3/50 or about 0.06
Oh my gosh, coffee break yall.
You guys are doing excellent! no worries!! We can do this! xD
screw the arithmatic, as long as you get the concepts
but I gotta turn this in ..hehe normal = -25x-0.06 ?
three tired brains is better than 1 tired brain i suppose - at least we catch everyone else's mistakes
you find a function for the slope (derivative), calculus over
finish with algebra
asfpaig the thing says both answers are wrong :(
graph them and see what it looks like
for the normal, same point and inverse negative slope, m = -25, point (1, 0.1), calculate that line again
not sure maybe all decimals or all fractions wanted, syntax
oo maybe. so how do i do 0.06 in fraction=.=
35/250? haha
normal y - 0.1 = -25*(x - 1)
y = 25.1 - 25x
y = -25x+25.1
haha yeah. So for the tangent it would be?
tan : 0.04x-0.06
Nope, it didn't like that.
AHA I GOT IT
oh dude, we got it at the same time lol. idk why i had the 0.06 negative.
...and nvm, I misread the problem xD
You guys are amazing <3 thanks for staying with me lol
yeah, dumb computer checking tools i guess you should choose decimal or fraction, and have to stay the same for everything
welcome, if you have another similar, i am sure it will be done in 2 or less responses, i promise
Right. Quick question though how come it had to be put in as 25.1-25x rather than -25x +25.1 ?
aw thank you:) i'm all done for today!! Though I should probably do some practice ones o.o clearly.
sorry was afk
i dont know, whoever programmed the thing did it
just always lead with a positive value if you can , i guess
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