Discover a wealth of knowledge and get your questions answered on IDNLearn.com. Our experts provide timely and accurate responses to help you navigate any topic or issue with confidence.
Sagot :
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].
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].
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Thank you for trusting IDNLearn.com with your questions. Visit us again for clear, concise, and accurate answers.
