Question

(a) Find \frac{d^2y}{dx^2} in terms of \(t\) for \(x = t^3 + 2t\), \(y = t^2\). \frac{d^2y}{dx^2} = \frac{2t(t^2 - 6)}{(t^2 + 2)^3} (b) Is the curve concave up or down at \(t = 1\)? At \(t = 1\), the curve is concave down

          (a) Find \frac{d^2y}{dx^2} in terms of \(t\) for \(x = t^3 + 2t\), \(y = t^2\).

\frac{d^2y}{dx^2} = \frac{2t(t^2 - 6)}{(t^2 + 2)^3}

(b) Is the curve concave up or down at \(t = 1\)?

At \(t = 1\), the curve is concave down

(a) Find (d^2y)/(dx^2) in terms of t for x = t^3 + 2t, y = t^2.

(d^2y)/(dx^2) = (2t(t^2 - 6))/((t^2 + 2)^3)

(b) Is the curve concave up or down at t = 1?

At t = 1, the curve is concave down

Added by Ashley K.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Eduard Sanchez

Step 1

Given that d^2y/dx^2 = 2t^6 - t^2 + 23x, we can substitute t = 1 into the equation: d^2y/dx^2 = 2(1)^6 - (1)^2 + 23x = 2 - 1 + 23x = 1 + 23x Now, we need to evaluate this expression at t = 1: d^2y/dx^2 = 1 + 23(1) = 1 + 23 Show more…

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