Rates Question1

Solution!

A.) Related Rates

B

a

b

c

12km

6km

?

90km/h

72km/h

A

c

c

@ t = 0, the diagram is shown

*The first step to solving a related rates problem would be to sketch a diagram of the situation to provide a "bigger picture" of the events

A.) Related Rates

a2 + b2 = c2 122 + 62 = c2144 + 36 = c2180 = c2

< pythagorean theorem relating distances

B

a

b

c

12km

6km

?

90km/h

72km/h

A

c

c

A.) Related Rates

a2 + b2 = c2 122 + 62 = c2144 + 36 = c2180 = c2

< pythagorean theorem relating distances

< exact value of the distance z

*With all three distances, we can now relate the three rates in the same manner

cB

a

b

c

12km

6km

?

90km/h

72km/h

A

c

c

A.) Related Rates

a2 + b2 = c2

< differentiate

*Differentiate each term with respect to time. Note that within each function, implicitly, there lies another function that represents rate.

B

a

b

c

12km

6km

?

90km/h

72km/h

A

c

c

A.) Related Rates

a2 + b2 = c2

< differentiate

*Input the distance between each car and the intersection at t = 0. Input the rates at which each car is moving with respect to the intersection.

B

a

b

c

12km

6km

?

90km/h

72km/h

A

c

c

A.) Related Rates

a2 + b2 = c2

< differentiate

*Now that we have the rate at which the distance between the cars A and B increases, we can move onto the next part of the question.

< solved through algebra; do not forget units!

B

a

b

c

12km

6km

?

90km/h

72km/h

A

c

c

B.) Antidifferentiation

s(t) = ys(0) = 4

*Like most questions, they may be intimidating but if you think thoroughly of what type of information is presented, the solution may be slightly clearer


s(t) = ys(0) = 4

< spot the variables, including the differentiating operator

*Note that the differentiable operator can be separated onto opposite sides of the equation, with different variables (y and t) on either sides. This creates a perfect opportunity to antidifferentiate.


s(t) = ys(0) = 4


< seperate the variables onto opposite sides of the equation

*As stated previously, a perfect opportunity to antidifferentiate both sides of the equation.


s(t) = ys(0) = 4


< separate the variables onto opposite sides of the equation

< antidifferentiate both sides


s(t) = ys(0) = 4



*ey has a nice antiderivative of itself while on the opposite side of the equation, a simple polynomial function is antidifferentiated. Remember that when antidifferentiated, "c" a constant, is produced that describes which function it is from the family of functions.


< combine both "c" constants to form a larger "C" constant on one side of the equation


s(t) = ys(0) = 4



*Now that we have "C", we have the complete s(t) function!


< combine both "c" constants to form a larger "C" constant on one side of the equation

e4 = C < find the value of "C" by substituting the coordinate (0,4): notice that all terms with the variable (t) are reduced to 054.5982 = C


*Remember that a Natural logarithm is an exponent with base e! Therefore we take the natural log of both sides to produce the function equal to y, or more specifically, s(t).

< function of s(t), with the addition of the value "C"


*In this step, we input the value retrieved from part A, as the time required in minutes for this remote controlled car to reach its destination.




s(112.6978) = ln(21440266.42) < algebra!



s(112.6978) = ln(21440266.42)s(112.6978) = 16.8808 km

< algebra!< solution rounded to four decimal places

The bicycle travelled a total distance of 16.8808 km.

Finish!

Rates Question1

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