To find [tex]\(\frac{dy}{dx}\)[/tex] given the parametric equations [tex]\(x = a t^2\)[/tex] and [tex]\(y = 2at\)[/tex], we'll follow a systematic approach.
1. Express [tex]\(x\)[/tex] and [tex]\(y\)[/tex] in terms of [tex]\(t\)[/tex]:
[tex]\[
x = a t^2
\][/tex]
[tex]\[
y = 2a t
\][/tex]
2. Calculate the derivatives [tex]\(\frac{dx}{dt}\)[/tex] and [tex]\(\frac{dy}{dt}\)[/tex]:
- Differentiating [tex]\(x\)[/tex] with respect to [tex]\(t\)[/tex]:
[tex]\[
\frac{dx}{dt} = \frac{d}{dt}(a t^2) = 2at
\][/tex]
- Differentiating [tex]\(y\)[/tex] with respect to [tex]\(t\)[/tex]:
[tex]\[
\frac{dy}{dt} = \frac{d}{dt}(2at) = 2a
\][/tex]
3. Find [tex]\(\frac{dy}{dx}\)[/tex] using the chain rule:
[tex]\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\][/tex]
4. Substitute the derivatives [tex]\(\frac{dy}{dt}\)[/tex] and [tex]\(\frac{dx}{dt}\)[/tex] into the equation:
[tex]\[
\frac{dy}{dx} = \frac{2a}{2at} = \frac{2a}{2at} = \frac{1}{t}
\][/tex]
Therefore, the answer is:
[tex]\[
\frac{dy}{dx} = \frac{1}{t}
\][/tex]
So, the correct choice is (a) [tex]\(\frac{1}{t}\)[/tex].
