![]() ![]() I hope that answers your question, and please let us know if not! : ) It’s only through experience (and heading down a lot of ultimately-unproductive paths) while we were learning this stuff ourselves that we now know more quickly which the best approach to take is. We’ve labeled the angle $\theta$ that the ladder makes with the ground, since the problem is asking us to find the rate at which that angle changes, $\dfrac$ as its own problem – see “ How fast is the ladder’s top sliding?” But then you’re solving an entirely different problem as another sub-part of this problem when it’s not necessary.)īut we want to emphasize that starting your solution attempt with $\tan \theta = y/x$ is perfectly reasonable. Draw a picture of the physical situation. To solve this problem, we will use our standard 4-step Related Rates Problem Solving Strategy.ġ. At what rate is the angle between the ladder and the ground changing when the base is 8 ft from the house? The base of the ladder starts to slide away from the house at 2 ft/s. There is no cost to you for having an account, other than our gentle request that you contribute what you can, if possible, to help us maintain and grow this site.At what rate does the angle change as a ladder slides away from a house?Ī 10-ft ladder leans against a house on flat ground. We believe that free, high-quality educational materials should be available to everyone working to learn well. You will also be able to post any Calculus questions that you have on our Forum, and we'll do our best to answer them! We do use aggregated data to help us see, for instance, where many students are having difficulty, so we know where to focus our efforts. ![]() Your selections are for your use only, and we do not share your specific data with anyone else. Your progress, and specifically which topics you have marked as complete for yourself.Your self-chosen confidence rating for each problem, so you know which to return to before an exam (super useful!).Your answers to multiple choice questions.Once you log in with your free account, the site will record and then be able to recall for you: (And it’s through our calculations that we find out exactly how dh/dt depends on h: as you see above, it turns out to depend on 1/h^2.) So we have to calculate the value of dh/dt for the moment the question tells us, which here happens to be when h = 10 cm. That is, dh/dt starts small and gets larger as the water level drops - it doesn’t stay the same throughout - which means that dh/dt _somehow_ depends on the value of h at that instant. Your physical sense of what’s going on here might help too: if you think about water draining out of a cone, you probably have the sense that at first (when the cone is full) the water level drops slowly, but then toward the end (when the cone is almost empty) the water level drops quickly. When it comes to substitute values, it’s important to look back and see what each variable means so that you don’t use a wrong value. ![]() The problem asks us to find dh/dt *at the instant when h = 10 cm*,” so that’s the value of h we have to use here.Ī key take-away is that that figure at the top matters a LOT: in a problem like this, it does a great deal of work in defining the variables for us and acts as a guide to everything that follows. Thanks for asking, Nicky! To answer it, look back at the figure at the top where we defined the different variables: “h” is the value of the water’s height in the cone at every instant, so as time passes and the water drains, the value of h changes.
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