Question

Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ at the given point without eliminating the parameter $x = \frac{1}{3}t^3 + 1$, $y = \frac{1}{4}t^4 + t$, $t = 3$. $\frac{dy}{dx} = \frac{t^3 + 1}{t^2}$ $\frac{d^2y}{dx^2} = \frac{t^3 + 1}{t(\frac{1}{3}t^3 + 1)}$

          Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ at the given point without eliminating the parameter
$x = \frac{1}{3}t^3 + 1$, $y = \frac{1}{4}t^4 + t$, $t = 3$.
$\frac{dy}{dx} = \frac{t^3 + 1}{t^2}$
$\frac{d^2y}{dx^2} = \frac{t^3 + 1}{t(\frac{1}{3}t^3 + 1)}$

Find (dy)/(dx) and (d^2y)/(dx^2) at the given point without eliminating the parameter
x = (1)/(3)t^3 + 1, y = (1)/(4)t^4 + t, t = 3.
(dy)/(dx) = (t^3 + 1)/(t^2)
(d^2y)/(dx^2) = (t^3 + 1)/(t((1)/(3)t^3 + 1))

Added by Vanesa N.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

11/07/2022

Step 1

dy/dt = d/dt((1/4)t^4 + t) = t^3 + 1 dx/dt = d/dt((1/3)t^3 + 1) = t^2 Show more…

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Find (dy)/(dx) and (d^(2)y)/(dx^(2)) at the given point without eliminating the parameter. x = (1/3)t^3 + 1, y = (1/4)t^4 + t, t = 3. (dy)/(dx) = (t^3 + 1)/(t^2) (d^(2)y)/(dx^(2)) = (t^3 + 1)/(t((1/3)t^3 + 1))