AI HL · AHL 2
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Related rates with the chain rule
AHL 2.5
A 6.5 m ladder leans on a wall. Its foot slides away at a steady 0.3 m/s. Drag the foot to choose the moment, and see how fast the top is falling then.
The ladder never changes length, so x² + y² = 6.5². Differentiating with respect to t and dividing by 2 gives x(dx/dt) + y(dy/dt) = 0. The arrows show the speeds: orange for the foot, teal for the top.
Drag the foot of the ladder.
If \(y=x^3+2x\), find \(\frac{dy}{dt}\) when \(x=3\) and \(\frac{dx}{dt}=4\).
Have a go first. Then open the working one line at a time.
\(\frac{dy}{dt}=\frac{dy}{dx}\times\frac{dx}{dt}\)
We know dx/dt and want dy/dt, so the chain rule joins them.
\(\frac{dy}{dx}=3x^2+2\)
Differentiate y with respect to x, as usual.
\(\frac{dy}{dt}=(3\times3^2+2)\times4=29\times4=116\)
Put in x = 3 and dx/dt = 4 only after differentiating.
y is increasing at 116 units per unit of time.
Positive means increasing.
If \(p=e^{-0.4q}\), find the rate of change of p when \(q=5\) and \(\frac{dq}{dt}=3\).
Have a go first. Then open the working one line at a time.
\(\frac{dp}{dt}=\frac{dp}{dq}\times\frac{dq}{dt}\)
Same link as before, with q in place of x.
\(\frac{dp}{dq}=-0.4e^{-0.4q}\)
The derivative of \(e^{kq}\) is \(ke^{kq}\).
\(\frac{dp}{dt}=-0.4e^{-2}\times3=-0.162\)
q = 5 gives \(-0.4\times5=-2\) in the power.
p is decreasing at 0.162 per unit of time.
The negative sign means p is falling.
If \(y=\ln(x+3)\), find the rate of increase of y when \(x=5\) and x is decreasing at 2 units per hour.
Have a go first. Then open the working one line at a time.
\(\frac{dx}{dt}=-2\)
Decreasing means the rate is negative.
\(\frac{dy}{dx}=\frac{1}{x+3}\)
The derivative of \(\ln(x+3)\) is \(\frac{1}{x+3}\).
\(\frac{dy}{dt}=\frac{1}{5+3}\times(-2)=-0.25\)
Substitute x = 5.
dy/dt is negative, so y is decreasing at 0.25 units per hour.
The question asked for the rate of increase. A negative answer is allowed.
Given that \(A=x^2+y^2\), find \(\frac{dA}{dt}\) when \(x=5\), \(y=12\), \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=-1\).
Have a go first. Then open the working one line at a time.
\(\frac{d}{dt}(x^2)=2x\frac{dx}{dt}\)
x changes with time, so the chain rule adds dx/dt.
\(\frac{dA}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}\)
The same step for \(y^2\).
\(\frac{dA}{dt}=2(5)(2)+2(12)(-1)=20-24=-4\)
Put the values in last.
A is decreasing at 4 square units per unit of time.
The negative term from y wins.
A rectangle has length x cm and width y cm. When \(x=5\) and \(y=3\), \(\frac{dx}{dt}=2\) cm s\(^{-1}\) and \(\frac{dy}{dt}=-0.5\) cm s\(^{-1}\). Find the rate of change of the area at this time.
Have a go first. Then open the working one line at a time.
\(A=xy\)
Both sides change, so differentiate a product.
\(\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}\)
Differentiate with respect to t.
\(\frac{dA}{dt}=5(-0.5)+3(2)=-2.5+6=3.5\)
Use the instant given.
The area is increasing at 3.5 cm² s⁻¹.
Include units.