To find the value of [tex]\(\frac{d f}{d t}\)[/tex] at [tex]\(t = 0\)[/tex] for the function [tex]\(f(x, y, z) = xy + z\)[/tex] where [tex]\(x = \cos t\)[/tex], [tex]\(y = \sin t\)[/tex], and [tex]\(z = t\)[/tex], we'll take the following steps:
1. Express [tex]\(f\)[/tex] in terms of [tex]\(t\)[/tex]:
Since [tex]\(x = \cos t\)[/tex], [tex]\(y = \sin t\)[/tex], and [tex]\(z = t\)[/tex], we can write:
[tex]\[
f(t) = (\cos t)(\sin t) + t
\][/tex]
Simplifying this, we get:
[tex]\[
f(t) = \cos t \sin t + t
\][/tex]
2. Differentiate [tex]\(f(t)\)[/tex] with respect to [tex]\(t\)[/tex]:
We'll use the product rule for differentiation and differentiate each term separately:
[tex]\[
\frac{df}{dt} = \frac{d}{dt}(\cos t \sin t) + \frac{d}{dt}(t)
\][/tex]
Let's differentiate the first term using the product rule [tex]\((uv)' = u'v + uv'\)[/tex] where [tex]\(u = \cos t\)[/tex] and [tex]\(v = \sin t\)[/tex]:
[tex]\[
\frac{d}{dt}(\cos t \sin t) = (\frac{d}{dt}(\cos t)) (\sin t) + (\cos t)(\frac{d}{dt}(\sin t))
\][/tex]
Knowing the derivatives of trigonometric functions, we have [tex]\(\frac{d}{dt}(\cos t) = -\sin t\)[/tex] and [tex]\(\frac{d}{dt}(\sin t) = \cos t\)[/tex]:
[tex]\[
\frac{d}{dt}(\cos t \sin t) = (-\sin t)(\sin t) + (\cos t)(\cos t)
\][/tex]
Simplifying this:
[tex]\[
\frac{d}{dt}(\cos t \sin t) = -\sin^2 t + \cos^2 t
\][/tex]
For the second term:
[tex]\[
\frac{d}{dt}(t) = 1
\][/tex]
Therefore, combining these results:
[tex]\[
\frac{df}{dt} = -\sin^2 t + \cos^2 t + 1
\][/tex]
3. Evaluate [tex]\(\frac{df}{dt}\)[/tex] at [tex]\(t = 0\)[/tex]:
Substituting [tex]\(t = 0\)[/tex] into the derivative obtained:
[tex]\[
\frac{df}{dt}\bigg|_{t=0} = -\sin^2(0) + \cos^2(0) + 1
\][/tex]
Knowing that [tex]\(\sin(0) = 0\)[/tex] and [tex]\(\cos(0) = 1\)[/tex]:
[tex]\[
\frac{df}{dt}\bigg|_{t=0} = -(0)^2 + (1)^2 + 1 = 0 + 1 + 1 = 2
\][/tex]
So, the value of [tex]\(\frac{df}{dt}\)[/tex] at [tex]\(t = 0\)[/tex] is [tex]\(\boxed{2}\)[/tex].
