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Tutorial Exercise Find $\frac{dx}{dt}$, $\frac{dy}{dt}$, and $\frac{dy}{dx}$. $x = 8t^3 + 6t$, $y = 9t - 7t^2$ Step 1 We are given the following parametric equations and asked to find derivatives with respect to t and also $\frac{dy}{dx}$. $x = 8t^3 + 6t$ $y = 9t - 7t^2$ Recall that x and y are simply functions of t, so we find their derivatives with respect to t as usual. $\frac{dx}{dt} = $ $\frac{dy}{dt} = $

          Tutorial Exercise
Find $\frac{dx}{dt}$, $\frac{dy}{dt}$, and $\frac{dy}{dx}$.
$x = 8t^3 + 6t$,
$y = 9t - 7t^2$
Step 1
We are given the following parametric equations and asked to find derivatives with respect to t and also $\frac{dy}{dx}$.
$x = 8t^3 + 6t$
$y = 9t - 7t^2$
Recall that x and y are simply functions of t, so we find their derivatives with respect to t as usual.
$\frac{dx}{dt} = $
$\frac{dy}{dt} = $

Tutorial Exercise
Find (dx)/(dt), (dy)/(dt), and (dy)/(dx).
x = 8t^3 + 6t,
y = 9t - 7t^2
Step 1
We are given the following parametric equations and asked to find derivatives with respect to t and also (dy)/(dx).
x = 8t^3 + 6t
y = 9t - 7t^2
Recall that x and y are simply functions of t, so we find their derivatives with respect to t as usual.
(dx)/(dt) =
(dy)/(dt) =

Added by Alexander W.

Calculus: Early Transcendentals

James Stewart 8th Edition

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Step 1: We are given the parametric equations x = 8t^3 + 6t and y = 9t - 7t^2. Show more…

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Tutorial Exercise Given the parametric equations x = 8t^3 + 6t and y = 9t - 7t^2, we are asked to find the derivatives with respect to t and also dx. Step 1: We are given the following parametric equations and asked to find derivatives with respect to t and also dx. x = 8t^3 + 6t y = 9t - 7t^2 Recall that x and y are simply functions of t, so we find their derivatives with respect to t as usual. dx/dt = 24t^2 + 6 dy/dt = 9 - 14t