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Shuuya

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I can't understand the solution, help please :')



Thanks!
 

InteGrand

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I can't understand the solution, help please :')



Thanks!
Note z is the distance between the two cars at any time t. Also, x is the distance of the car on the horizontal axis from the intersection, and y is the distance of the car on the vertical axis from the intersection.

The rate we are asked to find is dz/dt.

By Pythagoras, z^2 = x^2 + y^2.

The solutions then used implicit differentiation to find z*z' = x*x' + y*y', where a prime denotes differentiation wrt time.

Since at the instant in question, we are given x, x', y and y', we can find z at that instant and thus solve to find z' at that instant.
 

Shuuya

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Note z is the distance between the two cars at any time t. Also, x is the distance of the car on the horizontal axis from the intersection, and y is the distance of the car on the vertical axis from the intersection.

The rate we are asked to find is dz/dt.

By Pythagoras, z^2 = x^2 + y^2.

The solutions then used implicit differentiation to find z*z' = x*x' + y*y', where a prime denotes differentiation wrt time.

Since at the instant in question, we are given x, x', y and y', we can find z at that instant and thus solve to find z' at that instant.
Implicit differentiation isn't part of the 3U course, is it? :O
 

InteGrand

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I suppose a typical 3U method that completely avoids implicit differentiation (and is thus unnecessarily tedious) is to explicitly find x(t) and y(t) (as functions of t). They are: x(t) = 30 - 50t, and y(t) = 40 - 40t.

Then you can find z explicitly in terms of t by writing z(t) = sqrt((x(t))^2 + (y(t))^2) = … (insert expressions for x(t) and y(t) and simplify).

Then once you've z(t), you can calculate z'(t), since we have the expression z(t) available. Then sub. in t = 0 to find z'(0) (which is what we're after).

Implicit differentiation is really nothing more than the chain rule, so they should put it into 2U in my opinion (since 2U has chain rule).
 

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