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In implicit differentiation, what happens when you differentiate a y-term w.r.t. x?
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All Flashcards in Topic 5.14
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5.14.18 cards
In implicit differentiation, what happens when you differentiate a y-term w.r.t. x?
It picks up a factor of dy/dx (the chain rule), because y is treated as a function of x.
d/dx(y²) = ?
2y·dy/dx.
How do you differentiate a term like xy w.r.t. x?
Product rule: y + x·dy/dx.
After differentiating implicitly, how do you isolate dy/dx?
Collect all dy/dx terms on one side, factor out dy/dx, then divide.
dy/dx for x² + y² = 25?
2x + 2y·dy/dx = 0 ⇒ dy/dx = −x/y.
How do you find a gradient at a specific point on an implicit curve?
Substitute the point's x and y values into the expression for dy/dx.
When is the tangent to an implicit curve horizontal? Vertical?
Horizontal when the numerator of dy/dx is 0; vertical when the denominator is 0 (gradient undefined).
Equation of a tangent once you have m at (x₁, y₁)?
y − y₁ = m(x − x₁).
5.14.28 cards
What is the chain-rule link in a related-rates problem (V depending on r)?
dV/dt = dV/dr · dr/dt.
What are the three steps of a related-rates problem?
1) Write the formula linking the quantities. 2) Differentiate w.r.t. time t. 3) Substitute the known rate and value, then solve.
Which formula links the base and height of a sliding ladder?
Pythagoras: x² + y² = L² (L the fixed ladder length).
What does a negative rate of change mean?
The quantity is decreasing (e.g. the top of a ladder sliding down, or a tank draining).
Sphere: dV/dr = ? (V = (4/3)πr³)
dV/dr = 4πr² (the surface area).
For a cone with r = h/2, how do you simplify V = (1/3)πr²h before differentiating?
Substitute r = h/2 to get V = (1/3)π(h/2)²h = πh³/12, a function of h only.
d/dx of arctan(u)?
u′/(1 + u²).
When should you substitute the given numbers in a related-rates problem?
Last — differentiate the general formula first, then substitute the rate and value at that instant.
5.14.38 cards
What are the four steps of an optimisation problem?
1) Write the quantity. 2) Reduce to one variable using the constraint. 3) Differentiate and set = 0. 4) Justify max/min and answer the question.
What condition holds at the optimum value?
The first derivative is zero (a stationary point).
How do you justify a stationary point is a maximum?
Second-derivative test: f″ < 0 ⇒ maximum (or f′ changes + to − through the point).
How do you justify a stationary point is a minimum?
f″ > 0 ⇒ minimum (or f′ changes − to + through the point).
Why minimise D² instead of D for a shortest-distance problem?
D² has the same minimising x as D (it's a monotone transformation) but avoids the messy square-root derivative.
Open box from an a×a sheet (corner squares x): volume formula?
V = x(a − 2x)², with domain 0 < x < a/2.
Why must you check the domain / reject some solutions?
Lengths can't be negative and some roots make a dimension zero — those are physically impossible.
After finding the optimum variable, what is often still required?
The optimum VALUE — substitute the variable back into the original quantity.
Topic 5.14 study notes
Full notes & explanations for Implicit & related rates (HL only)
Math AA exam skills
Paper structures, command terms & tips
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