A particle moves in the xy-plane so that its position for t ≥ 0 is given by the parametric equations x = ln (t + 1) and y = k(t^2) , where k is a positive constant. The line tangent to the particle’s path at the point where t = 3 has slope 8. What is the value of k ?

Question

Create a detailed visual representation of the mathematical scenario described. Depict the xy-plane with both the x and y axes. Show a particle moving along a curvaceous path based on the parametric equations x = ln(t + 1) and y = k(t^2), with 't' being any positive number and 'k' being a positive constant which needs to be determined. At a particular point where t = 3, illustrate the line tangent to the particle’s path, with the slope of this tangent line being 8. Make sure to not include any text in the image.

A particle moves in the xy-plane so that its position for t ≥ 0 is given by the parametric equations x = ln (t + 1) and y = k(t^2) , where k is a positive constant. The line tangent to the particle’s path at the point where t = 3 has slope 8. What is the value of k ?

Answer 1

dy/dx = dy/dt / dx/dt = 2kt(t+1)

So, at t=3, dy/dx = 24k = 8
so, k = 1/3

Answer 2

Well, well, well, it seems like we've got ourselves a math problem! Don't worry, I'm here to help, armed with jokes and mathematical wisdom.

Let's find the slope of the tangent line by differentiating the parametric equations. Grab your calculus cap, my friend!

We have x = ln(t + 1) and y = k(t^2).

Differentiating x with respect to t, we get dx/dt = 1/(t + 1).

Differentiating y with respect to t, we get dy/dt = 2kt.

Now, we know that the slope of the tangent line at the point where t = 3 is 8. So, let's plug in t = 3 into our derivatives:

dx/dt = 1/(3 + 1) = 1/4.

dy/dt = 2k(3) = 6k.

Since the slope of the tangent line is 8, we can set up an equation:

8 = dy/dt ÷ dx/dt
8 = (6k)/(1/4)
8 = (6k)(4)
8 = 24k

Solving for k, we get k = 8/24 = 1/3.

So, my friend, the value of k is 1/3.

Remember, laughter is always the best constant in life, even in math problems! Keep smiling, my friend!

Answer 3

To find the value of k, we need to determine the slope of the tangent line at t = 3 using the given parametric equations for x and y.

First, let's find the derivative of y with respect to t to get the slope of the tangent line:

dy/dt = 2kt

To find the slope at t = 3, we substitute t = 3 into the derivative expression:

dy/dt = 2k(3) = 6k

Given that the slope of the tangent line at t = 3 is 8, we can set up an equation:

6k = 8

Now, solve for k:

k = 8/6

Simplifying the fraction:

k = 4/3

Therefore, the value of k is 4/3.

Answer 4

To find the value of k, we need to use the information given about the slope of the tangent line.

The slope of a tangent line to a curve given parametrically can be found using the chain rule. In this case, the slope is given as 8, so we need to find the derivative of y with respect to x and set it equal to 8.

First, let's find the derivative of y with respect to t:
dy/dt = 2kt

Next, we'll find the derivative of x with respect to t:
dx/dt = 1 / (t + 1)

Now, we can use the chain rule to find dy/dx:
dy/dx = (dy/dt) / (dx/dt)
= (2kt) / (1 / (t + 1))
= 2kt(t + 1)

We know that the slope dy/dx at t = 3 is equal to 8, so we can set up the equation:
8 = 2k(3)(3 + 1)

Simplifying the equation:
8 = 2k(12)
8 = 24k

Dividing both sides of the equation by 24, we find:
k = 8/24 = 1/3

Therefore, the value of k is 1/3.