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. In Equation ( 4.1.2) we see this is when cos 2 θ is largest this is when cos θ = 1, or when θ = 0.For that we would require to express height h as a function of time t. Common sense tells us this is when the car is directly in front of the camera (i.e., when θ = 0). We want to know the fastest the camera has to turn. Now take the derivative of both sides of Equation ( 4.1.1) using implicit differentiation: We need to convert the measurements so they use the same units rewrite -100 mph in terms of ft/s:ĭ x d t = - 100 m hr = - 100 m hr ⋅ 5280 ft m ⋅ 1 3600 hr s = - 146. Letting x represent the distance the car is from the point on the road directly in front of the camera, we haveĪs the car is moving at 100 mph, we have d x d t = - 100 mph (as in the last example, since x is getting smaller as the car travels, d x d t is negative). Figure 4.1.2 suggests we use a trigonometric equation.
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