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 ?
dy/dx = dy/dt / dx/dt = 2kt(t+1)
So, at t=3, dy/dx = 24k = 8
so, k = 1/3
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!
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.
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.