Answer:
The rate of change of the area formed by the ladder at that instant is 7 square metres per minute.
Step-by-step explanation:
Let x be the length of wall. And y be the distance of ladder from the wall.
Then, the decrement in the distance between the top of the ladder and the ground is,
[tex]\dfrac{dx}{dt}=-3[/tex]
Use Pythagoras theorem with base y and height x, and length as hypotaneous as,
[tex]h^{2}=y^{2}+x^{2}\\10^{2}=y^{2}+x^{2}\\100=y^{2}+x^{2}[/tex]
Differentiate the above equation with respect to time as,
[tex]100=y^{2}+x^{2}\\0=2\dfrac{dy}{dt} +2\dfrac{dx}{dt}\\-2y\dfrac{dy}{dt} =2x\dfrac{dx}{dt}\\-2y\dfrac{dy}{dt} =2x(-3)\\\dfrac{dy}{dt} =\dfrac{3x}{y}[/tex]
The area is given as,
[tex]A=\dfrac{1}{2}xy[/tex]
Differentiate to obtain the change in area as,
[tex]\dfrac{dA}{dt} =\dfrac{1}{2}y\dfrac{dx}{dt}+\dfrac{1}{2}x\dfrac{dy}{dt} \\\dfrac{dA}{dt} =\dfrac{1}{2}y\dfrac{dx}{dt}+\dfrac{1}{2}x\dfrac{dy}{dt}\\[/tex]
Here, y= 6 and x =8 by Pythagoras theorem. Substitute the values as,
[tex]\dfrac{dA}{dt} =\dfrac{1}{2}y\dfrac{dx}{dt}+\dfrac{1}{2}x\dfrac{dy}{dt}\\\dfrac{dA}{dt} =\dfrac{1}{2} \times 6 \times (-3)+\dfrac{1}{2} \times 8 \times \dfrac{3x}{y}\\\dfrac{dA}{dt} =\dfrac{1}{2} \times 6 \times (-3)+\dfrac{1}{2} \times 8 \times \dfrac{3 \times 8}{6}\\\dfrac{dA}{dt} = 7[/tex]
Thus, the rate of change of the area formed by the ladder at that instant is 7 square metres per minute.
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