Question

Help me on a question please!

Answer 1

i got derivative to be 1 / 2sqrtx

Answer 2

that is right

Answer 3

although, since you are asked for the derivative at \(x=t\) i guess you could write
\[\frac{1}{2\sqrt{t}}\]

Answer 4

why they bother changing one variable to another is beyond me

Answer 5

yes.. im not sure what do they mean :/

Answer 6

they want you to think of \(t\) as a number and \(x\) as a variable, so at \(x=t\) the slope is \(\frac{1}{2\sqrt{t}}\) and the equation for the line is
\[y-(\sqrt{t}-1)=\frac{1}{2\sqrt{t}}(x-t)\]

Answer 7

that is \(y-y_1=m(x-x_1)\) where \(x_1=t, y_1=\sqrt{t}-1\)

Answer 8

yes :)

Answer 9

if the tangent passes through the origin, then you can put \(x=0, y=0\) and solve for \(t\) in \[y-(\sqrt{t}-1)=\frac{1}{2\sqrt{t}}(x-t)\]

Answer 10

i think i have got it now
2y sqrt t - t + 2sqrt t - x = 0

Answer 11

i think if it passes through the origin you get \(t=4\)

Answer 12

i would leave the equation for the tangent in point - slope form as \[y-(\sqrt{t}-1)=\frac{1}{2\sqrt{t}}(x-t)\] or, if you want to solve for \(y\) put \[y=\frac{1}{2\sqrt{t}}(x-t)+(\sqrt{t}-1)\]

Answer 13

then simplify that one, if you want
it is really up to you

Answer 14

ahh ok thanks so much, i appreciate it

Answer 15

yw
check if i am right and that if it passes through the origin you will get \(t=4\)
makes a nicer line

Answer 16

yea the back of the book says t =4 :)