Use two ways of finding the slope of the tangent line explain why the up the dif

Question

Use two ways of finding the slope of the tangent line explain why the up the diffrential eqaution, using x and y, which gives the path (tractrix) of the weight, is dy/dx = - y/sqrt a^2 - y^2

Explanation / Answer

I don't know if you are allowed to use calculus in your class, so I will explain how to do each using calculus and not using calculus. First question: Using calculus: I'm sure you have learned that v=delta(x)/delta(t). If we look at the limit as t->0 for this, you see that we have v=dx/dt. Therefore, in order to plot a velocity-time graph from a position-time graph, you simply plot the derivative of the position-time graph. Not using calculus: Look at the graph you have and find the slope of the tangent lines at a number of the points. Plot the vaule of the slope for the same time, t, and put a best fit line to your points. Note that in this case, the velocity-time graph will be a zero-slope line, v=75 m/s (since the problem stated that the plane moves with constant speed +75 m/s). Second question: Using calculus: Take the integral from two to five of the velocity-time graph (assuming you are given an equation. If not, do it the "not using calculus" way). Not using calculus: Find the area under the velocity-time graph between times 2 s and 5 s, and that gives you the displacenet of the plane b/w those two times.

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