1 A particle is moving in a stright line with the position at any time t is given by s= t³–9t²+3t+1,where s in meters and t in second. Find its position and acceleration when velocity is ¯24m/s
2. Given that x²cos y_sin y=0,(0,π).
A. Verify that the given points on the curve.
B.use implicit differention to find the slope of the above curve at the given point.
C.find the equation of tangent and normal to the curve at that.
s = t^3-9t^2+3t+1
v = 3t^2-18t+3 = 3(t^2-6t+1)
so, solve for t when v=-24
a = 6t-18
then plug in that t to find v and a.
what is your problem with minus signs (-) ? over- and under-score really are ugly!
x^2 cosy - siny = 0
Surely you can plug in the point to verify it fits!
for dy/dx, just use the chain and product rules:
2x cosy - x^2 siny y' - cosy y' = 0
now just collect terms and solve for y':
y' = (2x cosy)/(x^2 siny + cosy)
or, if cosy≠0,
2x/(x^2 tany + 1)
1. To find the position and acceleration of the particle when the velocity is -24 m/s, we need to determine the values of time (t) and position (s) at that specific velocity.
First, let's find the velocity function by finding the derivative of the position function with respect to time, t.
s = t³ - 9t² + 3t + 1 (position function)
v = ds/dt = 3t² - 18t + 3 (velocity function)
Now, we set the velocity function equal to -24 m/s and solve for t:
-24 = 3t² - 18t + 3
Rearranging the equation:
3t² - 18t - 27 = 0
Next, we solve the quadratic equation. We can factor out a common factor of 3 from each term:
3(t² - 6t - 9) = 0
Now, we use either factoring or the quadratic formula to solve for t. In this case, let's use the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
Plugging in the values, a = 1, b = -6, and c = -9:
t = (-(-6) ± √((-6)² - 4 * 1 * (-9))) / (2 * 1)
t = (6 ± √(36 + 36)) / 2
t = (6 ± √(72)) / 2
t = (6 ± 6√2) / 2
t = 3 ± 3√2
Since time cannot be negative in this context, we can ignore the negative value:
t = 3 + 3√2
Now that we have the value of t, we can find the corresponding position (s) and acceleration (a).
To find the position (s), we substitute the value of t back into the position function:
s = (3 + 3√2)³ - 9(3 + 3√2)² + 3(3 + 3√2) + 1
s ≈ -74.99 m
To find the acceleration (a), we differentiate the velocity function with respect to time, t:
a = dv/dt = d²s/dt² = 6t - 18
Substituting the value of t:
a = 6(3 + 3√2) - 18
a ≈ -5.76 m/s²
Therefore, when the velocity is -24 m/s, the position of the particle is approximately -74.99 m and the acceleration is approximately -5.76 m/s².
2. A. To verify that the given points (0, π) lie on the curve of the equation x²cos(y) - sin(y) = 0, we substitute the values x = 0 and y = π into the equation:
(0)²cos(π) - sin(π) = 0
0 - 0 = 0
0 = 0
Since the equation is true, the given points (0, π) do lie on the curve.
B. To find the slope of the curve at the given point, we will use implicit differentiation.
Differentiate both sides of the equation with respect to x:
d/dx(x²cos(y) - sin(y)) = d/dx(0)
2xcos(y) - x²sin(y)dy/dx - cos(y)dy/dx = 0
Since we are finding the slope at the point (0, π), substitute x = 0 and y = π into the equation:
2(0)cos(π) - (0)²sin(π)dy/dx - cos(π)dy/dx = 0
-1 * dy/dx - (-1 * dy/dx) = 0
-2 * dy/dx = 0
Divide both sides by -2:
dy/dx = 0
The slope of the curve at the point (0, π) is 0.
C. To find the equation of the tangent and normal to the curve at the point (0, π), we will use the point-slope form of a line.
1. Tangent line: Since the slope (dy/dx) at the point (0, π) is 0, the tangent line will be a horizontal line passing through the point.
The equation of the tangent line is:
y - π = 0 * (x - 0)
y = π
Therefore, the equation of the tangent line to the curve at the point (0, π) is y = π.
2. Normal line: The normal line to a curve is perpendicular to the tangent line at a given point. Since the tangent line is horizontal, the normal line will be vertical.
The equation of the normal line is x = a, where a is the x-coordinate of the point.
Therefore, the equation of the normal line to the curve at the point (0, π) is x = 0.