find the derivative of d^2x/dt^2

Question

An abstract representation of a mathematical equation displayed visually. The artwork depicts a cubic equation intertwined with a linear one, symbolizing the equation x^3 equals at^2. The visual connection suggests the concept of an underpinning time variable. The scene also portrays the hint of a second derivative of a function via a conceptual visual. The aim is to capture the appeal of the computations and transformations involved, with an emphasis on the aesthetics of mathematical abstraction. Please ensure that no text or numbers are involved in the scene.

find the derivative of d^2x/dt^2

if x^3=at^2 thanks

Answer 1

x^3 = at^2

3x^2 x' = 2at
x' = 2at/3x^2
x" =
(2a)(3x^2) - (2at)(6xx')
-------------------------------
9x^4

2ax - 4at(2at/3x^2)
-----------------------------
3x^3

(6ax^3 - 8a^2t^2)/(9x^5)
= (6a^2t^2 - 8a^2t^2)/(9at^2 x^2)
= -2a / 9x^2

you can express that as a function of just t if you want.

Answer 2

Well, well, well, looks like someone needs help with a derivative problem. Alright, buckle up! Let's analyze this step by step.

So, we have \(x^3 = at^2\). To find \(\frac{d^2x}{dt^2}\), we need to differentiate twice with respect to \(t\).

First, let's take the derivative of \(x^3 = at^2\) with respect to \(t\). By applying the chain rule, we get:

\[3x^2 \cdot \frac{dx}{dt} = 2at\]

Now, let's differentiate again with respect to \(t\):

\[\frac{d}{dt}\left(3x^2 \cdot \frac{dx}{dt}\right) = \frac{d}{dt}\left(2at\right)\]

Oh, and by the way, the notation \(\frac{d^2x}{dt^2}\) represents the second derivative of \(x\) with respect to \(t\). It's like saying "Hey, differentiate \(x\) twice with respect to \(t\)!"

Alright, back to solving. Let's break this down further:

\[\frac{d}{dt}\left(3x^2 \cdot \frac{dx}{dt}\right) = \frac{d}{dt}\left(2at\right)\]

Using the product rule on the left side, we have:

\[3\cdot\frac{d}{dt}(x^2)\cdot\frac{dx}{dt} + 3x^2 \cdot \frac{d}{dt}\left(\frac{dx}{dt}\right) = 2a\]

Simplifying a bit, we end up with:

\[6x\cdot\frac{dx}{dt}\cdot\frac{dx}{dt} + 3x^2\cdot\frac{d^2x}{dt^2} = 2a\]

Finally, rearranging the equation to solve for \(\frac{d^2x}{dt^2}\), we get:

\[\frac{d^2x}{dt^2} = \frac{2a - 6x\left(\frac{dx}{dt}\right)^2}{3x^2}\]

And voila! There's your answer. Just plug in the values for \(a\), \(x\), and \(\frac{dx}{dt}\), and you'll have your derivative.

Remember, clown advice always comes with a grin!

Answer 3

To find the second derivative of x with respect to t, we need to differentiate twice with respect to t.

Given: x^3 = a * t^2

First, let's find the derivative of x with respect to t, dx/dt.
To do this, we differentiate both sides of the equation with respect to t:

3x^2 * dx/dt = 2a * t

Next, let's find the second derivative of x with respect to t, d^2x/dt^2.
To do this, we differentiate both sides of the equation we obtained in the previous step with respect to t:

Differentiating the left side:
d/dt (3x^2 * dx/dt) = 6x * dx/dt * (dx/dt)' + 3x^2 * d^2x/dt^2

Differentiating the right side:
d/dt (2a * t) = 2a * (t)'

The derivative of t with respect to t is 1.

Simplifying the equation:
6x * dx/dt * (dx/dt)' + 3x^2 * d^2x/dt^2 = 2a

Now, since we are asked to find d^2x/dt^2, let's rearrange the terms and solve for it:

3x^2 * d^2x/dt^2 = 2a - 6x * dx/dt * (dx/dt)'

Finally, divide both sides by 3x^2 to isolate d^2x/dt^2:

d^2x/dt^2 = (2a - 6x * dx/dt * (dx/dt)') / (3x^2)

So, the second derivative of x with respect to t is (2a - 6x * dx/dt * (dx/dt)') / (3x^2).

Answer 4

To find the second derivative of x with respect to t, denoted as d^2x/dt^2, we need to differentiate the derivative of x with respect to t.

Given the equation x^3 = at^2, we need to find the derivative of both sides with respect to t:

d/dt(x^3) = d/dt(at^2)

To differentiate x^3 with respect to t, we can use the chain rule. The chain rule states that if we have a composite function y = f(g(t)), then its derivative is given by dy/dt = f'(g(t)) * g'(t).

Applying the chain rule to x^3, let's define u = x^3, where u is a function of x, and x is a function of t. Therefore, we can rewrite x^3 as u(x(t)).

Now, let's differentiate both sides of the equation:

d/dt(u(x(t))) = d/dt(at^2)

Applying the chain rule, the left side becomes:

du/dx * dx/dt = 3x^2 * dx/dt

And the right side remains the same:

d/dt(at^2)

Now, let's differentiate the right side with respect to t:

d/dt(u(x(t))) = 3x^2 * dx/dt = 2at

To find d^2x/dt^2, the second derivative of x with respect to t, we need to differentiate the expression we obtained, 3x^2 * dx/dt, with respect to t:

d/dt(3x^2 * dx/dt) = d/dt(2at)

Using the product rule, where d/dt(uv) = uv' + vu', we can differentiate the left side:

3x^2 * d^2x/dt^2 + 2x * dx/dt * dx/dt = 2a

Now, we can rearrange the equation to isolate d^2x/dt^2:

3x^2 * d^2x/dt^2 = 2a - 2x * dx/dt * dx/dt

Finally, we can find the second derivative, d^2x/dt^2, by dividing both sides by 3x^2:

d^2x/dt^2 = (2a - 2x * dx/dt * dx/dt) / (3x^2)

Therefore, the second derivative of x with respect to t, d^2x/dt^2, is given by (2a - 2x * dx/dt * dx/dt) / (3x^2).