Question

Please help me on a, b, and c.

Answer 1

`\textsf{a)} \quad -(26)/(41)\; \sf cm/min`

`\textsf{b)} \quad -(258)/(1681)\; \sf deg/min`

`\textsf{c)} \quad 111\; \sf cm^2/min`

Explanation:

Define the variables:

• Let x = width of the rectangle
• Let y = length of the rectangle

Given that the width of the rectangle is decreasing at a rate of 2 cm/min:

`\implies \frac{\text{d}x}{\text{d}t}=-2\; \sf cm/min`

Given that the length of the rectangle is increasing at a rate of 6 cm/min:

`\implies \frac{\text{d}y}{\text{d}t}=6\; \sf cm/min`

Part (a)

The diagonal forms a right triangle.

Therefore, using Pythagoras Theorem:

`\implies x^2+y^2=D^2`

where D is the length of the diagonal.

To find the rate of change of the diagonal, take the derivative of the equation with respect to time (t):

`\implies \frac{\text{d}}{\text{d}t}x^2+\frac{\text{d}}{\text{d}t}y^2=D^2\frac{\text{d}}{\text{d}t}`

`\implies 2x \frac{\text{d}x}{\text{d}t}+2y\frac{\text{d}y}{\text{d}t}=2D\frac{\text{d}D}{\text{d}t}`

`\implies x \frac{\text{d}x}{\text{d}t}+y\frac{\text{d}y}{\text{d}t}=D\frac{\text{d}D}{\text{d}t}`

Given:

• y = 9 cm
• D = 41 cm

Use Pythagoras Theorem to calculate the width of the rectangle:

`\implies x^2+y^2=D^2`

`\implies x^2+9^2=41^2`

`\implies x=√(41^2-9^2)`

`\implies x=40\; \sf cm`

Given parameters:

• 
  `\frac{\text{d}x}{\text{d}t}=-2\; \sf cm/min`

• 
  `\frac{\text{d}y}{\text{d}t}=6\; \sf cm/min`
• 
  `x=40\;\sf cm`

• 
  `y = 9\; \sf cm`

• 
  `D=41\; \sf cm`

Substitute the given parameters into the equation:

`\implies x \frac{\text{d}x}{\text{d}t}+y\frac{\text{d}y}{\text{d}t}=D\frac{\text{d}D}{\text{d}t}`

`\implies (40) (-2)+(9)(6)=41\frac{\text{d}D}{\text{d}t}`

`\implies -26=41\frac{\text{d}D}{\text{d}t}`

`\implies \frac{\text{d}D}{\text{d}t}=-(26)/(41)`

Therefore, the length of the diagonal is decreasing at a rate of ²⁶/₄₁ cm/min.

Part (b)

Note: The angle θ has not been marked on the given diagram.

Therefore, I have used the angle BAC (please see attached diagram).

If θ is the angle BAC then to find the rate of change of the angle, find an expression for the angle using the tan ratio:

`\implies \tan \theta=(x)/(y)`

`\implies \tan \theta=xy^(-1)`

To find the rate of change of the angle, take the derivative of the equation with respect to time (t):

`\implies \frac{\text{d}}{\text{d}t}\tan \theta=\frac{\text{d}}{\text{d}t}xy^(-1)`

`\implies \sec^2 \theta \frac{\text{d}\theta}{\text{d}t}=(1)/(y)\frac{\text{d}x}{\text{d}t}-(x)/(y^2)\frac{\text{d}y}{\text{d}t}`

`\implies (1)/(\cos^2 \theta) \frac{\text{d}\theta}{\text{d}t}=(1)/(y)\frac{\text{d}x}{\text{d}t}-(x)/(y^2)\frac{\text{d}y}{\text{d}t}`

Given:

• x = 40 cm
• y = 9 cm
• D = 41 cm

Find an expression for the cosine of angle θ using the given parameters:

`\implies \cos \theta=(y)/(D)=(9)/(41)`

Given parameters:

• 
  `\frac{\text{d}x}{\text{d}t}=-2\; \sf cm/min`

• 
  `\frac{\text{d}y}{\text{d}t}=6\; \sf cm/min`
• 
  `x=40\;\sf cm`

• 
  `y = 9\; \sf cm`

• 
  `\cos \theta=(9)/(41)`

Substitute the given parameters into the equation:

`\implies (1)/(\cos^2 \theta) \frac{\text{d}\theta}{\text{d}t}=(1)/(y)\frac{\text{d}x}{\text{d}t}-(x)/(y^2)\frac{\text{d}y}{\text{d}t}`

`\implies \left((41)/(9)\right)^2 \frac{\text{d}\theta}{\text{d}t}=(1)/(9)(-2)-(40)/(9^2)(6)`

`\implies \left((1681)/(81)\right) \frac{\text{d}\theta}{\text{d}t}=-(2)/(9)-(80)/(27)`

`\implies\frac{\text{d}\theta}{\text{d}t}=-(86)/(27) \left((81)/(1681)\right)`

`\implies\frac{\text{d}\theta}{\text{d}t}=-(258)/(1681)\; \sf deg/min`

Therefore, the angle is decreasing at a rate of ²⁵⁶/₁₆₈₁ deg/min.

Part (c)

The equation for the area of triangle ABC is:

`A=(1)/(2)xy`

To find the rate of change of the area, take the derivative of the equation with respect to time (t):

`\implies \frac{\text{d}}{\text{d}t}A=\frac{\text{d}}{\text{d}t}(1)/(2)xy`

`\implies \frac{\text{dA}}{\text{d}t}=(1)/(2)y\frac{\text{d}x}{\text{d}t}+(1)/(2)x\frac{\text{d}y}{\text{d}t}`

Given parameters:

• 
  `\frac{\text{d}x}{\text{d}t}=-2\; \sf cm/min`

• 
  `\frac{\text{d}y}{\text{d}t}=6\; \sf cm/min`
• 
  `x=40\;\sf cm`

• 
  `y = 9\; \sf cm`

Substitute the given parameters into the equation:

`\implies \frac{\text{dA}}{\text{d}t}=(1)/(2)y\frac{\text{d}x}{\text{d}t}+(1)/(2)x\frac{\text{d}y}{\text{d}t}`

`\implies \frac{\text{dA}}{\text{d}t}=(1)/(2)(9)(-2)+(1)/(2)(40)(6)`

`\implies \frac{\text{dA}}{\text{d}t}=-9+120`

`\implies \frac{\text{dA}}{\text{d}t}=111\; \sf cm^2/min`

Therefore, the area of the triangle is increasing at a rate of 111 cm²/min.