Question

find an equation for the tangent line to the graph of: f(x)=1/2sec(x)-cos(x)
at the point (1/3p, f(1/3p))
[p means pi]

Answer 1

first find the derivative of the function.

Answer 2

okay I got f'(x)=1/2tanxsecx+sinx
what do I do now?

Answer 3

wait. is it
\[ \large f(x)=\frac{1}{2\sec x}-\cos x \]
??

Answer 4

no (1/2)secx-cosx
my b

Answer 5

\[ \large f(x)=\frac{1}{2}\sec x-\cos x \] then
\[ \large f'(x)=\frac{1}{2}\sec x\tan x+\sin x \]
right?

Answer 6

yes so where do I go from there?

Answer 7

first compute
\[ \large f(1/3\pi)= \]

Answer 8

is it 1/2?

Answer 9

\[ \large f(1/3\pi)=1/2\sec(1/3\pi)-\cos(1/3\pi)=1/2 \]

Answer 10

then u have the point \((1/3\pi,1/2)\)
the tangent line has the form
\[ \large y-f(1/3\pi)=f'(1/3\pi)(x-1/3\pi) \]
\[ \large y-1/2=f'(1/3\pi)(x-1/3\pi) \]

Answer 11

so \[fprime(1/3\pi)=\sqrt{3}+(\sqrt{3}+4)\] ?

Answer 12

u have to compute
\[ \large f'(1/3\pi)= \]

Answer 13

I did I was just wondering if I did it right

Answer 14

i got 2.5980

Answer 15

okay yeah I have to keep it in radical form but I think those answers are the same thing.

Answer 16

\[ \large f'(1/3\pi)=1/2\sec(1/3\pi)\tan(1/3\pi)+\sin(1/3\pi) \]
\[ \large =1/2\cdot 2\cdot\sqrt{3}+\frac{\sqrt{3}}{2}=3/2\sqrt{3} \]

Answer 17

I don't see where you got that answer

Answer 18

\[ \large \sin1/3\pi=\sqrt{3}/2 \]
and
\[ \large \cos1/3\pi=1/2 \]
isn't it??

Answer 19

yes but I'm not seeing how you got \[3/2\sqrt{3}\]

Answer 20

well
\[ \large =\sqrt{3}+\frac{\sqrt{3}}{2}=\sqrt{3}(1+1/2)=3/2\cdot\sqrt{3} \]

Answer 21

ohhh ok got it. so the final answer should be \[2y-3\sqrt{3}x=1-\sqrt{3}\pi\]

Answer 22

yes

Answer 23

ok thanks! do you have time to help me answer one more problem?

Answer 24

yes why not
i hope it not too long

Answer 25

alright thanks. here it is:
a particle moves along a line so that at a time, t, where \[0\le t \le \pi,\] its position is given by \[s(t)=-4\cos t-(t^2/2)+10\]. find the velocity of the particle when its acceleration is zero.

Answer 26

velocity if s' and acceleration is s''
right?

Answer 27

*is not if

Answer 28

yes that's right. how would I take the derivative of that equation? its kinda intimidating

Answer 29

then compute \(s'(t)\) and \(s''(t)\):
\[ \large s'(t)=4\sin t-t \]
and
\[ \large s''(t)=4\cos t-1 \]

Answer 30

intimidating?? the one u did first was way harder

Answer 31

now solve s''(x)=0

Answer 32

both are intimidating. yeah I found this help site off of a website that is for k-8th grade but I'm taking ap calc. okay thanks that helped a lot. so when you plug o in for s"(x) then you get 1.3181. then you plug that t value into the velocity equation. right?

Answer 33

yes

Answer 34

what is K8th??

Answer 35

kindergarten thru 8th grade

Answer 36

so u r like 15??

Answer 37

keep it up!!

Answer 38

haha thanks!

Answer 39

u r welcome
glad to help