wild deratee (2)(1)
TRANSCRIPT
QUEST to BECOME A MATHEMON MASTER!Wild Deratee Appears!: Related Rates Question
.:. THE SITUATION .:.
After finally carrying out their plan of attack, Jane and Derivee find themselves in a tight situation as they take on the wild Deratee. Although their hearts burn with determination, they’re about to find out that Deratee is a master of the fearsome Calculus art known as Related Rates!
QUESTION…BATTLEDeratee used QUESTION!
Deratee was 40 km north of some stranger at 9 am. After defeating the stranger, Deratee begins moving west at 10 km/h and the stranger begins running, weakly, eastwards at 30 km/h. At 3 pm, how fast is the distance between the stranger and Deratee changing?
Derivee is overwhelmed! Derivee takes 16 damage!
SOLUTION…BATTLE
Derivee recovered 20 HP!
Jane used a DIAGRAM!
COUNTER ATTACK!
Deratee used GLARE!
Derivee’s DEF has lowered!
SOLUTION…BATTLE
Derivee recovered 20 HP!
Jane used a DIAGRAM SHUFFLE!
SOLUTION…BATTLE
Deratee used PROVOKE!
Derivee is unaffected!!!
SOLUTION…BATTLEDerivee used LIST!
This is a list of information that we are given from the question. This helps significantly in related rates questions. In this question, take note that:
The Pythagorean theorem can be applied;
Deratee takes 7 damage!
22240 xDS
2 2 240 ( )S D x
10dD
dt 30
dS
dt
COUNTER ATTACK!
Deratee used GROWL!
Derivee flinched! Derivee’s DEF has lowered!
SOLUTION…BATTLEDerivee used IMPLICIT DIFFERENTIATION!
We simply differentiate both sides of the previous function we just made implicitly
Remember the derivative of a constant is zero and also the chain rule must be applied since there is a function within a function
2( )( ) 2 ( )dS dD dx
S D xdt dt dt
Its super effective! Deratee takes 24 damage! Deratee is paralyzed!
SOLUTION…BATTLE
This is just a simplified form of the function previously done. We isolate dz/dt because we are looking for the change in distance of the stranger and Deratee.
Deratee takes 5 damage! Deratee is paralyzed!
( )( )dS dD
S D dzdt dtz dt
Derivee used REARRANGE!
SOLUTION…BATTLE
To find the change in distance we’ll need to know the values of the variables S and D.
Since we are given the change of rate of both S and D we can calculate distance.
Using unit analysis we see that we will need to multiply the rate by a value of time to end up with a unit of distance
The question states that Deratee defeated the stranger at 9 am and now we are finding its distance at 3pm. The 6 hour time that has passed is what we’ll use to multiply to both rates.
(30 )(6 )km
S hh
180km
10 (6 )km
D hh
60km
Derivee used FIND DISTANCES!
Now that we know the values of S and D we can plug them into our equation we made earlier from using the Pythagorean Theorem
All the parts are now known so that we can solve for x, which is the distance between Deratee and the stranger.
However, the question is asking for the rate that distance is changing.
2 2 240 ( )S D x
2 2 240 (180 60) x
59200km x
Deratee takes a critical hit! Deratee takes 16 damage!
COUNTER ATTACK!
Deratee recovered! Deratee used TACKLE!
Derivee takes 8 damage!
SOLUTION…BATTLE
dz/dt can now be solved because we know all the variables needed
Plugging in all the values completes the question!
( )( )dS dD
S D dzdt dtz dt
(180 60)(30 10)
59200
dz
dt
39.456km dz
h dt
Derivee used SUBSTITUTION!
ITS SUPER EFFECTIVE! Deratee takes 27 damage!
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........Gotcha! Deratee was caught!
Jane used a MATHECUBE!