Find dy dx for the parametric equations x equals one third t cubed and y =
-t^2 + t. (10 points)
A) -2t + 1/ t^2
B) negative 1 over t squared
C) negative 1 over t cubed
D) 2
x = (1/3)t^3 and y = -t^2 + t
dx/dt = t^2 and dy/dt = -2t + 1
dy/dx = (dy/dt)/(dx/dt) = (-2t + 1)/t^2
= -2t/t^2 + 1/t^2
= -2/t + 1/t^2
Unless you have a typo in A), none of the choices are correct.
To find \(\frac{{dy}}{{dx}}\) for the given parametric equations, we can use the chain rule.
The chain rule states that if \(x\) and \(y\) are both functions of a variable \(t\), then
\(\frac{{dy}}{{dx}} = \frac{{\frac{{dy}}{{dt}}}}{{\frac{{dx}}{{dt}}}}\).
So, first, let's find \(\frac{{dy}}{{dt}}\) and \(\frac{{dx}}{{dt}}\) separately.
Given: \(x = \frac{1}{3}t^3\) and \(y = -t^2 + t\)
To find \(\frac{{dx}}{{dt}}\):
Differentiate \(x\) with respect to \(t\):
\(\frac{{dx}}{{dt}} = \frac{{d}}{{dt}}(\frac{1}{3}t^3)\)
Applying the power rule for differentiation, we get:
\(\frac{{dx}}{{dt}} = \frac{1}{3} \cdot 3t^2\)
\(\frac{{dx}}{{dt}} = t^2\)
Now, let's find \(\frac{{dy}}{{dt}}\):
Differentiate \(y\) with respect to \(t\):
\(\frac{{dy}}{{dt}} = \frac{{d}}{{dt}}(-t^2 + t)\)
Using the sum rule for differentiation, we get:
\(\frac{{dy}}{{dt}} = \frac{{d}}{{dt}}(-t^2) + \frac{{d}}{{dt}}(t)\)
Applying the power rule for differentiation, we get:
\(\frac{{dy}}{{dt}} = -2t + 1\)
Now, we can find \(\frac{{dy}}{{dx}}\) by dividing \(\frac{{dy}}{{dt}}\) by \(\frac{{dx}}{{dt}}\):
\(\frac{{dy}}{{dx}} = \frac{{-2t + 1}}{{t^2}}\)
Therefore, the correct answer is A) \(-2t + \frac{{1}}{{t^2}}\)