Find a tangent vector at the given value of t for the following curves.

$r(t)=\langle e^t,e^{3t},e^{5t}\rangle,~t=0$

Question

Find a tangent vector at the given value of t for the following curves.

$r(t)=\langle e^t,e^{3t},e^{5t}\rangle,~t=0$

austinl · Answer

@amistre64 
Care to lend a helping hand my friend?

amistre64 · Answer

the derivative of a vector function is the derivative of its vector parts.

amistre64 · Answer

e^t to e^t
e^3t to 3e^t
e^5t to 5 e^5t

at t=0 we get;  1,3,5

but the tangent vector itself may need to be of unit length, so dividing it by its magnitude should suffice

austinl · Answer

$\large{r^{\prime}(t)=\langle e^t,3e^{3t},5e^{5t}\rangle}$

Correct?

Then,

$\large{|r^{\prime}(t)|=\sqrt{(e^t)^2+(3e^{3t})^2+(5e^{5t})^2}}$

Correct so far?

And $\large{T(t)=\frac{r^{\prime}(t)}{|r^{\prime}(t)|}}$

Is this correct?

amistre64 · Answer

correct so far yes;  but i see no real need to make it a unit vector since the directions ask for "a" tangent vector.  Not that it matters if you do or not :)

amistre64 · Answer

one thing I am mulling over is if we need to position it accordingly:

x(t) + k x'(t)
y(t) + k y'(t)
z(t) + k z'(t)

but that may be immaterial

austinl · Answer

I'm a bit fuzzy on all of this stuff.... so any input you have would be appreciated.

amistre64 · Answer

your first response is adequate enough.  it is a tangent vector, of unit length.

austinl · Answer

Then when I put it in fraction form, and simplify a little I assume, do I just plug in t=0?

amistre64 · Answer

yep

austinl · Answer

Alright, sounds good!

amistre64 · Answer

good luck ;)