Related Rates

Just so you know, related rates is actually the Application of Implicit Differentiation by using Chain Rule in the form of dy/dx = dy/du * du/dx.

Btw, at Khan academy it’s called the Differentiate related functions.

Refer to Khan lecture.
Refer to video by KristaKingMath: Related rates
Refer to video by The Organic Chemistry Tutor: Introduction to Related Rates

Strategy:

• Abstract every given condition algebraically, e.g. s(t), s'(t), V(t), V'(t)...
• Find a proper equation to CONNECT all these given conditions
• Differentiate both sides of equation, with using the Chain rule.
• Take back the given values to the Differentiated equation to get the result.

Example: Change of volumes

Refer to previous note of Implicit Differentiation.

Solve:

• Write out all conditions algebraically:
• r'(t)=1
• r(t)=5
• h'(t)=-4
• h(t)=8
• V(t) = π·r²·h
• So they’re asking for V'(t), it then becomes V'(t) = π·d/dt (r²·h)
• Apply Product rule & Chain rule then substitute: V'(t) = π((r²)'·h + r²h') = π(2r'·r + r²·h') = -20π

Notice that: (r²)' is a composite function, so you want to apply the Chain rule, =2r·r'

Example: Change of volumes

Solve:

• From the given conditions, we got that r'(t)=-12, r(t)=40 and h=2.5.
• We know the equation of cylinder’s volume is: V= π·[r(t)]²·h
• Differentiate both side of the equation to get:

• Take back all the known values into the equation get V'(t)=-2400π

Example: Change of area

Solve:

• Write out all conditions algebraically:
• d₁'(t) = -7
• d₁(t) = 4
• d₂'(t) = 10
• d₂(t) = 6
• A(t) = (d₁·d₂)/2
• So it’s asking for instantaneous rate of change, then it is A'(t) = (d₁'·d₂ + d₁·d₂')/2
• Apply the Product rule only, to get A'(t) = -1.

Example: Pythagorean Theorem

Refer to Khan academy lecture: Related rates: Approaching cars

Solve:

• It’s a problem to calculate two objects’ dynamic absolute distance: it’s getting closer and closer until distance become 0, which means they crashes at 0.
• Write down all the conditions algebraically:
• c₁'(t) = -10
• c₁(t) = 4
• c₂'(t) = -6
• c₂(t) = 3
• D(t) = √(c₁²+c₂²)
• So D'(t) = d/dt · (√(c₁²+c₂²)) = (2c₁'c₁+2c₂'c₂)/(2√(c₁²+c₂²))

Example: Sliding ladder

Example: multiple composite

Solve:

• Write down all the conditions algebraically:
• S'(t) = π/2
• S(t) = 12π
• S = 2π·r, which means r = S/2π
• A(t) = π·r²
• So A'(t) = d/dt · (π·r²) = 2π·r·r' = 2π·6·(S/2π)' = 6S' = 3π