Find all t such that r'(t) is perpendicular to r(t)

Let r(t) =
t E (all real numbers)

-What I know how to do: Steps
(1) (r)*(r'(t))

r(t) =
r'(t) = <3t^2, 1, 3>
(r)*(r'(t)) = (t^3)*(3t^2) + (t+1)*(1) + (3t)*(3) = 3t^5 + t + 1 + 9t
0 = 3t^5 + 10t + 1

(2) Since the function f(t) = 3t^5 + 10t + 1 is strictly increasing: check the derivative

At t= 0 (function is positive)
At t= -1 (function is negative)

Anwser: t ~ -0.1
*I am lost on step 2, I don't know how to get this answer.

Since f(t) is strictly increasing, f(t) = 0 at one point only.
This occurs between t = −1 and t = 0 since
f(−1) = −12 < 0
f(0) = 1 > 0

Next step: try t = −0.5
f(−0.5) = −4.09375 < 0 ----> t is between −0.5 and 0

Try t = −0.2
f(−0.2) = −1.00096 < 0 ----> t is between −0.2 and 0

Try t = −0.1
f(−0.1) = −0.0003 < 0 ----> t is between −0.1 and 0

Now we see that t = −0.1 gives result that is closer to 0 than t = 0 does.
So, to 1 decimal point, t = −0.1

Now we could use similar method to find t to more decimal places.
To 6 decimal places, we get t = −0.099997, so −0.1 is a pretty good approximation.

-----------------------------

We could also use Newton's method to find roots of a function.
http://en.wikipedia.org/wiki/Newton%27s_…

f(t) = 3t⁵ + 10t + 1
f'(t) = 15t⁴ + 10

Starting with initial estimate t₀, we can find better estimate using formula:
t₁ = t₀ − f(t₀)/f'(t₀)

Starting with t₀ = 0, we get:

t₁ = 0 − f(0)/f'(0) = 0 − 1/10 = −1/10
t₂ = −1/10 − f(-1/10)/f'(−1/10) = −0.099997
t₃ = −0.099997 − f(−0.099997)/f'(−0.099997) = −0.099997

Since third result is the same as second result, then we will not get anything more accurate to 6 decimal points.

t = −0.099997 ≈ −0.1

Wow, you're so bad @ Math.