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VARVARA [1.3K]
3 years ago
11

A hypothetical square grows so that the length of it's diagonals are increasing at a rate of 8 m/min. How fast is the area of th

e square increasing when the diagonals are 4m each?
Mathematics
1 answer:
guapka [62]3 years ago
5 0

Answer:

32m^{2}/min

Step-by-step explanation:

We let the length of the square be l, the diagonal be z and the area be A.

Then by Pythagoras theorem;

l^{2}+l^{2}=z^{2}\\2l^{2}=z^{2}

The are of a square of length l is given as;

A=l^{2}

Combining these two equations yields;

z^{2}=2A

A=\frac{z^{2} }{2}

Next, we have been given;

\frac{dz}{dt}=8

We are required to find;

\frac{dA}{dt} when z = 4

By chain rule;

\frac{dA}{dt}=\frac{dA}{dz}*\frac{dz}{dt}

Differentiating A=\frac{z^{2} }{2} with respect to z yields;

\frac{dA}{dz}=z

Therefore,

\frac{dA}{dt}=\frac{dA}{dz}*\frac{dz}{dt} = 8z

When z is 4m the area of the square will be increasing at a rate of;

8m/min * 4m = 32m^{2}/min

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