Problem #2
Given the diagram of the statement, we have:
From the diagram, we see that we have two triangles:
Triangle 1 or △ADP, with:
• angle ,θ,,
,• hypotenuse ,h = AP,,
,• adjacent cathetus, ac = AD = x cm.
,• opposite cathetus ,oc = DP,.
Triangle 2 or △OZP, with:
• angle θ,
,• hypotenuse, h = OP = 4 cm,,
,• adjacent cathetus, ac = ZP = AP/2,.
(a) △ADP: sides and area
Formula 1) From geometry, we know that for right triangles Pitagoras Theorem states:
[tex]h^2=ac^2+oc^2.[/tex]Where h is the hypotenuse, ac is the adjacent cathetus and oc is the opposite cathetus.
Formula 2) From trigonometry, we have the following trigonometric relation for right triangles:
[tex]\cos \theta=\frac{ac}{h}.[/tex]Where:
• θ is the angle,
,• h is the hypotenuse,
,• ac is the adjacent cathetus.
(1) Replacing the data of Triangle 1 in Formulas 1 and 2, we have:
[tex]\begin{gathered} AP^2=AD^2+DP^2\Rightarrow DP=\sqrt[]{AP^2-AD^2}=\sqrt[]{AP^2-x^2\cdot cm^2}\text{.} \\ \cos \theta=\frac{AD}{AP}=\frac{x\cdot cm}{AP}\text{.} \end{gathered}[/tex](2) Replacing the data of Triangle 2 in Formula 2, we have:
[tex]\cos \theta=\frac{ZP}{OP}=\frac{AP/2}{4cm}.[/tex](3) Equalling the right side of the equations with cos θ in (1) and (2), we get:
[tex]\frac{x\cdot cm}{AP}=\frac{AP/2}{4cm}.[/tex]Solving for AP², we get:
[tex]\begin{gathered} x\cdot cm=\frac{AP^2}{8cm}, \\ AP^2=8x\cdot cm^2\text{.} \end{gathered}[/tex](4) Replacing the expression of AP² in the equation for DP in (1), we have the equation for side DP in terms of x:
[tex]DP^{}=\sqrt[]{8x\cdot cm^2-x^2\cdot cm^2}=\sqrt[]{x\cdot(8-x)}\cdot cm\text{.}[/tex](ii) The area of a triangle is given by:
[tex]S=\frac{1}{2}\cdot base\cdot height.[/tex]In the case of triangle △ADP, we have:
• base = DP,
,• height = AD.
Replacing the values of DP and AD in the formula for S, we get:
[tex]S=\frac{1}{2}\cdot DP\cdot AD=\frac{1}{2}\cdot(\sqrt[]{x\cdot(8-x)}\cdot cm)\cdot(x\cdot cm)=\frac{x}{2}\cdot\sqrt[]{x\cdot(8-x)}\cdot cm^2.[/tex](b) Maximum value of S
We must find the maximum value of S in terms of x. To do that, we compute the first derivative of S(x):
[tex]\begin{gathered} S^{\prime}(x)=\frac{dS}{dx}=\frac{1}{2}\cdot\sqrt[]{x\cdot(8-x)}\cdot cm^2+\frac{x}{2}\cdot\frac{1}{2}\cdot\frac{8-2x}{\sqrt{x\cdot(8-x)}}\cdot cm^2 \\ =\frac{1}{2}\cdot\sqrt[]{x\cdot(8-x)}\cdot cm^2+\frac{x}{2}\cdot\frac{(4-x^{})}{\cdot\sqrt[]{x\cdot(8-x)}}\cdot cm^2 \\ =\frac{1}{2}\cdot\frac{x\cdot(8-x)+x\cdot(4-x)}{\sqrt[]{x\cdot(8-x)}}\cdot cm^2 \\ =\frac{x\cdot(6-x)}{\sqrt[]{x\cdot(8-x)}}\cdot cm^2\text{.} \end{gathered}[/tex]Now, we equal to zero the last equation and solve for x, we get:
[tex]S^{\prime}(x)=\frac{x\cdot(6-x)}{\sqrt[]{x\cdot(8-x)}}\cdot cm^2=0\Rightarrow x=6.[/tex]We have found that the value x = 6 maximizes the area S(x). Replacing x = 6 in S(x), we get the maximum area:
[tex]S(6)=\frac{6}{2}\cdot\sqrt[]{6\cdot(8-6)}\cdot cm^2=3\cdot\sqrt[]{12}\cdot cm^2=6\cdot\sqrt[]{3}\cdot cm^2.[/tex](c) Rate of change
We know that the length AD = x cm decreases at a rate of 1/√3 cm/s, so we have:
[tex]\frac{d(AD)}{dt}=\frac{d(x\cdot cm)}{dt}=\frac{dx}{dt}\cdot cm=-\frac{1}{\sqrt[]{3}}\cdot\frac{cm}{s}\Rightarrow\frac{dx}{dt}=-\frac{1}{\sqrt[]{3}}\cdot\frac{1}{s}\text{.}[/tex]The rate of change of the area S(x) is given by:
[tex]\frac{dS}{dt}=\frac{dS}{dx}\cdot\frac{dx}{dt}\text{.}[/tex]Where we have applied the chain rule for differentiation.
Replacing the expression obtained in (b) for dS/dx and the result obtained for dx/dt, we get:
[tex]\frac{dS}{dt}(x)=(\frac{x\cdot(6-x)}{\sqrt[]{x\cdot(8-x)}}\cdot cm^2\text{)}\cdot(-\frac{1}{\sqrt[]{3}}\cdot\frac{1}{s}\text{)}[/tex]Finally, we evaluate the last expression for x = 2, we get:
[tex]\frac{dS}{dt}(2)=(\frac{2\cdot(6-2)}{\sqrt[]{2\cdot(8-2)}}\cdot cm^2\text{)}\cdot(-\frac{1}{\sqrt[]{3}}\cdot\frac{1}{s})=-\frac{8}{\sqrt[]{12}}\cdot\frac{1}{\sqrt[]{3}}\cdot\frac{cm^2}{s}=-\frac{8}{\sqrt[]{36}}\cdot\frac{cm^2}{s}=-\frac{8}{6}\cdot\frac{cm^2}{s}=-\frac{4}{3}\cdot\frac{cm^2}{s}.[/tex]So the rate of change of the area of △ADP is -4/3 cm²/s.
Answers
(a)
• (i), Side DP in terms of x:
[tex]DP(x)=\sqrt[]{x\cdot(8-x)}\cdot cm\text{.}[/tex]• (ii), Area of ADP in terms of x:
[tex]S(x)=\frac{x}{2}\cdot\sqrt[]{x\cdot(8-x)}\cdot cm^2.[/tex](b) The maximum value of S is 6√3 cm².
(c) The rate of change of the area of △ADP is -4/3 cm²/s when x = 2.
