a 13-ft ladder is leaning against a wall. suppose that the base of the ladder sides away from the wall at the constant rate of 3ft/sec
1. explain why the motion of the two ends of the ladder can be represented by the parametric equations:
x1(t)=3t, y2(t)=0
x2(t)=0, y2(t)= radical(13^2-(3t)^2)
2. what values of t make sense in this problem situation?
3. use stimultaneous mode. give a viewing window that shows the action.(Hint: it may be helpful to "hide" th ecoordinate axes if your grapher has this feature. if not, adjust the parametric equation to move th eaction away from the axes)
4. use analytic method to find the rate at which the top of the ladder is moving down the wall at t=0.5, 1, 1.5, and 2sec. in theory, how fast is the top of the ladder moving as it hit the ground?
help me!
1. To represent the motion of the two ends of the ladder using parametric equations, we consider the ladder as a right-angled triangle. Let's assume that the bottom left corner of the wall is the origin of our coordinate system.
The bottom end of the ladder is sliding away from the wall at a constant rate of 3 ft/sec. This means that the x-coordinate of the bottom end, denoted as x1(t), increases by 3t ft as time t passes. Hence, x1(t) = 3t.
Since the bottom end slides along the ground, its y-coordinate remains constant at 0.
For the top end of the ladder, we need to consider the Pythagorean theorem. The ladder has a length of 13 ft. Let's denote the distance from the bottom end to the top end as y2(t). Then, according to the Pythagorean theorem, y2(t) can be calculated as the square root of the difference between the length of the ladder squared (13^2) and the square of the x-coordinate of the bottom end (3t)^2:
y2(t) = sqrt(13^2 - (3t)^2)
2. In this problem situation, we need to consider the values of t for which the ladder is still leaning against the wall. Since the ladder has a length of 13 ft, the values of t that make sense should satisfy the condition:
0 ≤ t ≤ 13/3
This ensures that the ladder is not completely extended or retracted, which would not be a realistic situation.
3. Since the question asks for the graphing of parametric equations, we can use a graphing calculator or software to visualize the motion of the ladder. For the viewing window, we can set the x-axis from 0 to approximately 5, and the y-axis from 0 to 13. This will allow us to observe the ladder sliding away from the wall within a reasonable range.
4. To find the rate at which the top of the ladder is moving down the wall at specific times, we can use the derivative of the y2(t) equation with respect to time. The derivative will represent the rate of change of the y-coordinate with respect to time at any given moment.
Taking the derivative of y2(t), we have:
dy2/dt = d/dt(sqrt(13^2 - (3t)^2))
= (-6t) / (2sqrt(13^2 - (3t)^2))
= -3t / sqrt(13^2 - (3t)^2)
To find the rate at times t = 0.5, 1, 1.5, and 2 sec, we substitute these values into the derivative equation and calculate the corresponding rates.
To determine the theoretical speed at which the top of the ladder is moving as it hits the ground, we need to find the value of t when y2(t) equals zero. By setting y2(t) equal to zero and solving for t, we can find the time at which the ladder touches the ground.