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IB Mathematics: Applications and Interpretation HL · AHL 2: Further Derivatives · AHL 2.5
Related rates of change Problem set A Worked solutions
Skills
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If \(y=2x^3-3x\), find \(\frac{dy}{dt}\) when \(x=2\) and \(\frac{dx}{dt}=3\).
- \(\frac{dy}{dx}=6 x^{2} - 3\)
- \(\frac{dy}{dt}=(6 x^{2} - 3)\frac{dx}{dt}\)
- \(=(21)(3)=63\)
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If \(y=e^{4z}\), find the rate of increase of y when \(z=0.5\) and \(\frac{dz}{dt}=3\).
- \(\frac{dy}{dz}=4e^{4z}\)
- \(\frac{dy}{dt}=4e^{2}\times3\)
- \(=88.7\)
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For each part, find the rate of increase of a.
(a) \(a=\frac{1}{b}\), when \(b=4\) and b is increasing at 2 per second.
- \(a=b^{-1}\)
- \(\frac{da}{db}=-b^{-2}=-\frac{1}{b^2}\)
- \(\frac{da}{dt}=-\frac{1}{4^2}\times2=-0.125\)
(b) \(a=\frac{1}{b^2}\), when \(b=2\) and b is increasing at 3 per second.
- \(a=b^{-2}\)
- \(\frac{da}{db}=-2b^{-3}=-\frac{2}{b^3}\)
- \(\frac{da}{dt}=-\frac{2}{2^3}\times3=-0.75\)
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If \(y=\ln(x+2)\), find the rate of increase of y when \(x=4\) and x is decreasing at 5 units per hour.
- \(\frac{dx}{dt}=-5\)
- \(\frac{dy}{dx}=\frac{1}{x+2}\)
- \(\frac{dy}{dt}=\frac{1}{6}\times(-5)=-5/6 = -0.833\)
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Given that \(B=x^3+y^3\), find \(\frac{dB}{dt}\) when \(x=2\), \(y=1\), \(\frac{dx}{dt}=3\) and \(\frac{dy}{dt}=-4\).
- \(\frac{dB}{dt}=3x^2\frac{dx}{dt}+3y^2\frac{dy}{dt}\)
- \(=3(2)^2(3)+3(1)^2(-4)\)
- \(=24\)
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Given that \(C=\frac{x}{y}\), find \(\frac{dC}{dt}\) when \(x=6\), \(y=2\), \(\frac{dx}{dt}=-1\) and \(\frac{dy}{dt}=3\).
- \(\frac{dC}{dt}=\frac{y\frac{dx}{dt}-x\frac{dy}{dt}}{y^2}\)
- \(=\frac{(2)(-1)-(6)(3)}{2^2}\)
- \(=-5\)
In context
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A sheet of dough is rolled out as a rectangle. At one moment its length is 30 cm and its width is 20 cm. The length is increasing at 3 cm s\(^{-1}\) and the width is decreasing at 0.5 cm s\(^{-1}\). Find the rate of change of the area at this moment, and say whether the area is increasing or decreasing.
- \(A=xy\)
- \(\frac{dA}{dt}=x\frac{dy}{dt}+y\frac{dx}{dt}\)
- \(=(30)(-0.5)+(20)(3)=45\)
- The area is increasing at 45 cm² s⁻¹.
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Atmospheric pressure P kPa at a height of h km is modelled by \(P=101e^{-0.12h}\). A weather balloon rises at 0.3 km per minute. Find the rate of change of the pressure when \(h=5\).
- \(\frac{dP}{dh}=-12.12e^{-0.12h}\)
- \(\frac{dP}{dt}=\frac{dP}{dh}\times\frac{dh}{dt}=-12.12e^{-0.6}\times0.3\)
- \(=-2.00\)
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The brightness of a lamp at a distance d metres is \(I=\frac{80}{d^2}\) lux. A student walks away from the lamp at 0.5 m s\(^{-1}\). Find the rate of change of the brightness when \(d=4\).
- \(I=80d^{-2}\)
- \(\frac{dI}{dd}=-160d^{-3}=-\frac{160}{d^3}\)
- \(\frac{dI}{dt}=-\frac{160}{4^3}\times0.5=-1.25\)
Stretch
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Two square plots have sides x m and y m. Their total area \(A=x^2+y^2\) stays constant while x is increasing at 2 m per day. Find the rate of change of y when \(x=6\) and \(y=8\).
- \(A=x^2+y^2\)
- \(\frac{dA}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\)
- \(2(6)(2)+2(8)\frac{dy}{dt}=0\)
- \(\frac{dy}{dt}=-\frac{24}{16}=-1.5\)
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An image on a screen is x cm wide and y cm high, and the ratio \(C=\frac{x}{y}\) stays constant as it is resized. When \(x=16\) and \(y=9\), the width is increasing at 2 cm s\(^{-1}\). Find the rate of change of the height.
- \(C=\frac{x}{y}\)
- \(\frac{dC}{dt}=\frac{y\frac{dx}{dt}-x\frac{dy}{dt}}{y^2}=0\)
- \((9)(2)-(16)\frac{dy}{dt}=0\)
- \(\frac{dy}{dt}=\frac{18}{16}=1.125\)
jmaths.xyz · AI HL · AHL 2.5 Related rates with the chain rule