a particle moves along the x axis in such a way that its acceleration at time t, t>0 , is given by x(t)= (ln x)^2. at what value of t does the velocity of the particle attain its maximum
To find the value of t when the velocity of the particle attains its maximum, we need to find the derivative of the velocity function and set it equal to zero.
Let's start by finding the velocity function. The velocity is the integral of the acceleration function with respect to t:
v(t) = ∫ (ln x)^2 dt
To evaluate this integral, we need to find an expression for x in terms of t. However, the expression for x is not given in the problem. Therefore, we cannot directly find the velocity function.
Hence, we cannot determine the value of t when the velocity of the particle attains its maximum without additional information about the relation between x and t.
To find the value of t at which the velocity of the particle attains its maximum, we need to follow these steps:
Step 1: Find the expression for the velocity of the particle.
Step 2: Identify the critical points of the velocity function by finding where its derivative is equal to zero.
Step 3: Analyze the critical points to determine which one corresponds to the maximum velocity.
Let's begin with Step 1:
The acceleration function is given as x(t) = (ln x)^2. However, we need to find the expression for velocity, not the acceleration. To do this, we need to integrate the acceleration function with respect to t.
∫x(t) dt = ∫(ln x)^2 dt
To solve this integral, we can make a substitution where u = ln x and du = (1 / x) dx. Rearranging, we have dx = x du. Therefore, the integral becomes:
∫(ln x)^2 dt = ∫u^2 x du
Integrating x with respect to t gives:
x(t) = ∫u^2 x du
Now we move on to Step 2:
To find the critical points, we take the derivative of x(t) with respect to t and set it equal to zero:
d/dt [x(t)] = d/dt [∫u^2 x du] = 0
Step 3:
By analyzing the critical points, we can determine which one represents the maximum velocity. To do this, we can evaluate the sign of the second derivative of x(t), which will tell us whether the point is a maximum or minimum.
Once you have the expression for the second derivative, evaluate it at the critical points found in Step 2. If the second derivative is positive, the critical point corresponds to a minimum velocity. However, if the second derivative is negative, the critical point corresponds to a maximum velocity.
By following these steps, you can find the value of t at which the velocity of the particle attains its maximum.
velocity has max when acceleration is zero
I assume x(t) = (ln t)^2 which is zero at t=1.
Assuming the function is
x(t)=(ln(t))^2
then t has a vertical asymptote at x=0.
which explains why the domain of x(t) is set at x>0.
To find the maximum, we need to find
x'(t)=0
where
x'(t)=2ln(t)/t
and x'(t)=0 at t=1.
Check if maximum or minimum by finding x"(t)=2/t^2 - 2ln(t)/t
so
x"(1)=2 >0
Hence t=1 is a minimum.
Therefore, the function has no maximum, as it increases indefinitely as t inreases for t>1.