Derivitive of Dot product

[steller]

let \(\displaystyle u(t) =< -\sqrt{t}sint, t, t^{2/3} \) and \(\displaystyle v(t) = < -\sqrt{t}sint, cos^2t, -t^{1/3}>\)

Compute \(\displaystyle {d/dt} (u(t) . v(t))\)

Taking the dot product i have:

\(\displaystyle tsint^2(t) + tcos^2(t) - t = 0\)

From here i think i see a trig identity but not sure what to do with the t in front of cos, sin.

So, we will proceed with another route!

Remove whats common t . The negative 1 comes from -t from moving whats common.
\(\displaystyle t(cos^2(t) - 1 + sin^2(t))=0\)

So now i have

\(\displaystyle t=0\) and \(\displaystyle cos^2(t)-1 + sin^2(t) = 0\)

wolfram said that this was an identity but i could not verify that!

\(\displaystyle cos^2(t)-1 = 1-sin^2(t)\)

plugging in the identity for cos^2(t) we have

\(\displaystyle 1-sin^2(t)-1 + sin^2(t) = 0\)

Therefore, \(\displaystyle t=0 \) and \(\displaystyle 0=0 \)

making the derivative 0?

However, i never took the derivative and i couldn't verify the identity.

Right after i take the dot product could i move the -t to the right and divide out the other t?
then use the cos^2 + sin^2 = 1 identity? then take the derivative?

Where did i go wrong?

 

[HallsofIvy]

let \(\displaystyle u(t) =< -\sqrt{t}sint, t, t^{2/3} \) and \(\displaystyle v(t) = < -\sqrt{t}sint, cos^2t, -t^{1/3}>\)

Compute \(\displaystyle {d/dt} (u(t) . v(t))\)

Taking the dot product i have:

\(\displaystyle tsint^2(t) + tcos^2(t) - t = 0\)

From here i think i see a trig identity but not sure what to do with the t in front of cos, sin.

Yes, you can factor out the "t" to get \(\displaystyle t(sin^2(t)+ cos^2(t))- t= t- t= 0\) for all t.

So, we will proceed with another route!

Remove whats common t . The negative 1 comes from -t from moving whats common.
\(\displaystyle t(cos^2(t) - 1 + sin^2(t))=0\)

That is, as I said before, identically equal to 0, but since you say you did not know that, where did you get the "0" on the right?

So now i have

\(\displaystyle t=0\) and \(\displaystyle cos^2(t)-1 + sin^2(t) = 0\)

wolfram said that this was an identity but i could not verify that!

\(\displaystyle cos^2(t)-1 = 1-sin^2(t)\)

plugging in the identity for cos^2(t) we have

\(\displaystyle 1-sin^2(t)-1 + sin^2(t) = 0\)

Therefore, \(\displaystyle t=0 \) and \(\displaystyle 0=0 \)

making the derivative 0?

However, i never took the derivative and i couldn't verify the identity.

\(\displaystyle sin^2(t)+ cos^2(t)= 1\)!

Right after i take the dot product could i move the -t to the right and divide out the other t?
then use the cos^2 + sin^2 = 1 identity? then take the derivative?

Where did i go wrong?

Since that dot product is identically 0, its derivative is identically 0.

 

[steller]

Thank you i think i see it now.