Now, to solve, take the derivative with respect to time of both sides, giving you:Īnd then doing the regular arithmetic operations you get: In other words, the constant area of the rectangle acts as a constraint because:ġ.) We know something about the Area (namely, that it remains constant)Ģ.) Both x & y are related to area via the formula Area = x*y However, if I tell you that the area of the rectangle is constant then there is enough information for you to determine the rate of change of side "y" with respect to time. The answer is that there is simply not enough information-"y" may be growing or shrinking or even remaining the same length. Also, you know that side "x" is growing 1 foot every minute. Let's say you have a rectangle and you know that side "x" is 3 feet long and side "y" is 2 feet long. PLEASE comment if you see error in logic or elsewhere. Also, time at that moment is like frozen, but don't want to get philosophical. And decreasing angle means decreasing sine of that angle over time. I think key thing to understand here is that adjacent side changes over time, that is making angle do change(decrease in our case) over time. That moment means that instant, what is time at an instant? Well we think it's infinitesimally close to zero, so we substitute in derivative t=0:ġ0*cos( arccos(8/10) ) * -1/sqrt( 1-(8/10)^2 ) *4/10 = 8 * -4/6 = -16/3 Įxatly at that moment we know that adjacent side is 8 and velocity is 4ft* time. So we multiply whole expression by 10 because argument of arrcos() function is making things in unit circle proportion and we have triangle that ladder is forming 10 times bigger.ġ0*sine(arccos(8+4t/10)) = Model of our eventĭ/dt = 10*d/dt =ġ0* cos( arccos(8+4t/10) ) * d/dt =ġ0*cos( arccos(8+4t/10) ) * -1/sqrt( 1-(8+4t/10)^2 ) * d/dt =ġ0*cos( arccos(8+4t/10 ) ) * -1/sqrt( 1-(8+4t/10)^2 ) * 4/10. Also we see that our hypotenuse is 10 times bigger that radius of unit circle. So, arccos(8+4t/10), here 8+4t represents change of adjacent side(at that moment the value is 8) over time and 10 represents constant lenght of hypotenuse. That is a nice thing so we can use arccos() function to find angle at that moment so we could plug it in sine function(angle decreases over time, sine decreases). We see that whatever change occures hypotenuse is constant. At the same time your angle is decreasing so is your sine, that is the thing you want find rate of in respect to time. You see that as time changes you cosine proportion (8/10 right at this moment) changes, it is getting bigger. Here is how I managed to solve it in different way. But that would definitely require modifying the problem (adding data), not just adding another question. If just for fun, you could make some reasonable assumptions about the mass and materials in question say that it started to slide with no initial velocity and try to express h(t). We can't figure the acceleration, because there is no information about the mass of the ladder and the materials that make up the ladder and the wall. The ladder has forces (gravitation and friction with the wall and the floor most importantly) acting on it, hence it should have acceleration (if those forces don't balance out). Using this approximation would assume that the rate of change of the height (at THAT MOMENT) stays the same or, in other words, there is no acceleration. We have figured out dh/dt, which is an approximation of our formula. However, we lack information to produce the formula needed. If we did this, then we just plug h=0 into the formula and solve for t. For that we would require to express height h as a function of time t.
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