. The acceleration of a particle moving back and forth on a line is
d².s/d.t² .π²cosπ.t m/s² at any time t. If s=0 and v=8m/s when t=0 ,find s when t=1sec..
2.find the equation for the tangent line to the curve at the point defind by ralue of t.also find d²y/dx² at the point x=sec.t,y= tan.t,wheret=π/4..
Sure, I can help you with both questions. Let's start with the first one.
1. Find s when t = 1 sec:
Given information:
Acceleration, a = d²s/dt² = π²cos(πt) m/s²
Initial velocity, v = 8 m/s
Initial position, s = 0
To solve this problem, we'll need to integrate the acceleration, a, twice with respect to time to find the position, s.
Step 1: Integration of acceleration to find velocity:
v = ∫ a dt
v = ∫ π²cos(πt) dt
v = (π²sin(πt))/π + C₁
v = πsin(πt) + C₁
Using the initial condition v = 8 m/s when t = 0:
8 = πsin(0) + C₁
C₁ = 8
So, the velocity equation becomes:
v = πsin(πt) + 8
Step 2: Integration of velocity to find position:
s = ∫ v dt
s = ∫ (πsin(πt) + 8) dt
s = (-πcos(πt))/π + 8t + C₂
s = -cos(πt) + 8t + C₂
Using the initial condition s = 0 when t = 0:
0 = -cos(0) + 0 + C₂
C₂ = 1
So, the position equation becomes:
s = -cos(πt) + 8t + 1
To find s when t = 1 sec:
s = -cos(π(1)) + 8(1) + 1
s = -cos(π) + 8 + 1
s = -(-1) + 9
s = 1 + 9
s = 10 meters
Therefore, the position of the particle when t = 1 sec is 10 meters.
Now, let's move on to the second question.
2. Find the tangent line equation and second derivative:
Given information:
x = sec(t)
y = tan(t)
t = π/4
To find the equation of the tangent line at the point defined by t = π/4, we need to find the derivative of y with respect to x and evaluate it at x = sec(t), y = tan(t), and t = π/4.
Step 1: Find dy/dx:
dy/dx = (dy/dt)/(dx/dt)
dx/dt = d(sec(t))/dt = sec(t)tan(t)
dy/dt = d(tan(t))/dt = sec²(t)
Therefore,
dy/dx = (sec²(t))/(sec(t)tan(t))
dy/dx = tan(t)
Step 2: Evaluate dy/dx at t = π/4:
dy/dx = tan(π/4)
dy/dx = 1
So, the slope of the tangent line at t = π/4 is 1.
Step 3: Find the equation of the tangent line:
We know that the equation of a line in point-slope form is y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line, and m is the slope.
Using the point (sec(t), tan(t)) and the slope 1, we have:
y - tan(t) = 1(x - sec(t))
Substituting t = π/4:
y - tan(π/4) = 1(x - sec(π/4))
y - 1 = x - √2
The equation of the tangent line at the point defined by t = π/4 is:
y = x - √2 + 1
Lastly, let's find d²y/dx² at x = sec(t), y = tan(t), where t = π/4.
d²y/dx² represents the second derivative of y with respect to x.
Step 4: Find d²y/dx²:
d²y/dx² = d(dy/dx)/dx
Taking the derivative of dy/dx = tan(t) with respect to x:
d(tan(t))/dx = sec²(t) * d(sec(t))/dx
Since dx/dt = sec(t)tan(t):
d(tan(t))/dx = sec²(t) * (dx/dt)²
Therefore,
d²y/dx² = sec²(t) * (dx/dt)²
Substituting t = π/4 and dx/dt = sec(t)tan(t):
d²y/dx² = sec²(π/4) * (sec(π/4)tan(π/4))²
sec(π/4) = √2:
d²y/dx² = (√2)² * (√2 * √2)²
d²y/dx² = 2 * 2 * 2
d²y/dx² = 8
Therefore, d²y/dx² at x = sec(t), y = tan(t), where t = π/4, is 8.
To find the position of the particle when t = 1 second, we can integrate the acceleration function twice with respect to time.
Given:
Acceleration, a = d²s/dt² = π²cos(πt) m/s²
First, let's integrate the acceleration function to find the velocity, v:
∫a dt = ∫π²cos(πt) dt
Integrating cos(πt) gives us sin(πt)/π:
v = ∫a dt = ∫π² cos(πt) dt = (π² sin(πt))/π = π sin(πt)
Next, let's find the constant of integration using the given initial condition:
When t = 0, s = 0 and v = 8 m/s.
Integrating v with respect to time gives us the position function, s:
s = ∫v dt = ∫π sin(πt) dt
Integrating sin(πt) gives us -cos(πt)/π:
s = ∫π sin(πt) dt = (-cos(πt))/π + C
Applying the initial condition s = 0 when t = 0:
0 = (-cos(π*0))/π + C
0 = (-1)/π + C
C = 1/π
Now we can find s when t = 1 second:
s = (-cos(πt))/π + C
s = (-cos(π))/π + 1/π
s = (1+1)/π
s = 2/π
So, the position of the particle when t = 1 second is 2/π meters.
Now, let's move on to the second question:
To find the equation for the tangent line to the curve at a specific point, we need both the derivative dy/dx and the values of x and y at the given point.
Given: x = sec(t), y = tan(t), where t = π/4.
First, let's find dy/dx:
dy/dx = (dy/dt)/(dx/dt)
Using the chain rule:
dy/dx = (dy/dt) * (dt/dx)
Here, dy/dt can be found by differentiating y = tan(t) with respect to t:
dy/dt = sec²(t)
Now, dx/dt can be found by differentiating x = sec(t) with respect to t:
dx/dt = sec(t) * tan(t)
Finally, substitute the given values t=π/4 into dy/dt, dx/dt, x, and y:
dy/dx = (dy/dt) * (dt/dx) = sec²(π/4) * (sec(π/4) * tan(π/4))
= (1/cos²(π/4)) * (1/cos(π/4) * sin(π/4))
= (1/1) * (1/1 * (1/√2))
= 1/√2
So, the derivative dy/dx at the point x = sec(t), y = tan(t), where t = π/4, is 1/√2.
To find the equation of the tangent line, we need the point and the slope. We already have the slope from dy/dx, and the point (sec(t), tan(t)).
Using the point-slope form of the equation of a line, the tangent line is:
y - tan(t) = (1/√2)(x - sec(t))
For the given values t = π/4, substitute sec(t) = sec(π/4) = √2 and tan(t) = tan(π/4) = 1:
y - 1 = (1/√2)(x - √2)
This is the equation for the tangent line at the point defined by the values t = π/4, x = sec(t), and y = tan(t).
Let's also find d²y/dx² at the point x = sec(t), y = tan(t), where t = π/4:
To find d²y/dx², we need to differentiate dy/dx with respect to x:
d²y/dx² = d(dy/dx)/dx
Using the quotient rule:
d²y/dx² = (d(dy/dx)/dt) * (dt/dx)
Again, dy/dx is 1/√2, which we derived earlier.
Now, dt/dx can be found by differentiating x = sec(t) with respect to x:
dt/dx = 1/(dx/dt)
dt/dx = 1/(sec(t) * tan(t))
Substituting the given values t = π/4:
dt/dx = 1/(sec(π/4) * tan(π/4))
= 1/(√2 * 1/√2)
= 1
So, dt/dx = 1.
Now, multiply the derived value of dy/dx with dt/dx:
d²y/dx² = (1/√2) * 1 = 1/√2
Therefore, d²y/dx² at the point x = sec(t), y = tan(t), where t = π/4 is 1/√2.