In Larry Gonick's book, "The Cartoon Guide to Calculus" (which shows you the level I'm at with the whole subject), he presents the problem of trying to calculate the horizontal/ground-velocity rate with which a plane is approaching a ground-based radar site. The plane is flying in a flat, level path toward the radar station, maintaining a constant 3 km altitude. The radar station sees that the plane's air-to-radar-site rate is -320 km/h. This, of course, sets up a right-triangle situation, with the air-to-radar-station rate (-320 km/h) being the hypotenuse, a rate which is related to the horizontal/ground rate at which the plane is approaching the radar station. All of this then allows the use of the Pythagorean theorem.
So far, this all seems straight-forward, but I am not doing something correctly in my attempts to solve the problem. Please show me where my thinking is awry. Here is my flawed attempt:
Let a = the horizontal velocity which we are trying to find, so its derivative is a' km/h. (This would be the horizontal, bottom leg of the triangle)
Let b = the constant 3km altitude of the plane, so it's derivative is b', which equals 0 km/h, since it is constant. (This is the vertical leg of the triangle)
Let c = the downward-slanting, air-to-radar-station rate of the plane's approach to the radar station, so its derivative, c' = -320 km/h (This is the triangle's hypotenuse)
1) Using the Pythagorean theorem to relate all of the above, we have c^2 = a^2 + b^2.
2) Since the derivative of each term is related to the other terms' derivatives by the above equation, we get: (c^2)' = (a^2)' + (b^2)' (or is this wrong?)
3) Because we want to solve for a', we transpose the terms of the equation, giving us: (a^2)' = (c^2)' - (b^2)'
4) Since, by the chain rule, (a^2)' = (2a)(a'); (c^2)' = (2c)(c'); and (b^2)' = (2b)(b'), then plugging in the given values for each terms, we get...
4) (2a)(1) = (2)(-320)(1) + (2)(3)(0), giving 2a = -640 + 0
5) Finally, solving for 'a,' we get a = -320 km/hr. ...which is wrong! Clearly, the rate of the plane-to-radar-station can't equal the rate of the plane's travel over the ground in the horizontal direction. The correct answer is -400 km/h.
So where did I screw up? Any help would be much appreciated.
So far, this all seems straight-forward, but I am not doing something correctly in my attempts to solve the problem. Please show me where my thinking is awry. Here is my flawed attempt:
Let a = the horizontal velocity which we are trying to find, so its derivative is a' km/h. (This would be the horizontal, bottom leg of the triangle)
Let b = the constant 3km altitude of the plane, so it's derivative is b', which equals 0 km/h, since it is constant. (This is the vertical leg of the triangle)
Let c = the downward-slanting, air-to-radar-station rate of the plane's approach to the radar station, so its derivative, c' = -320 km/h (This is the triangle's hypotenuse)
1) Using the Pythagorean theorem to relate all of the above, we have c^2 = a^2 + b^2.
2) Since the derivative of each term is related to the other terms' derivatives by the above equation, we get: (c^2)' = (a^2)' + (b^2)' (or is this wrong?)
3) Because we want to solve for a', we transpose the terms of the equation, giving us: (a^2)' = (c^2)' - (b^2)'
4) Since, by the chain rule, (a^2)' = (2a)(a'); (c^2)' = (2c)(c'); and (b^2)' = (2b)(b'), then plugging in the given values for each terms, we get...
4) (2a)(1) = (2)(-320)(1) + (2)(3)(0), giving 2a = -640 + 0
5) Finally, solving for 'a,' we get a = -320 km/hr. ...which is wrong! Clearly, the rate of the plane-to-radar-station can't equal the rate of the plane's travel over the ground in the horizontal direction. The correct answer is -400 km/h.
So where did I screw up? Any help would be much appreciated.
