The length of a rectangle is increasing at a rate of 6 cm/s and its width is increasing at a rate of 4 cm/s. when the length is 9 cm and the width is 4 cm, how fast is the area of the rectangle increasing?

Given :

To find :-

Solution :-

As we know that :-

A = lb

To find the rate :-

d(A)/dt = d(lb)/dt .

Differenciate :-

dA/dt = l (db/dt ) + b (dl/dt )

Substitute :-

dA/dt = 9*4 + 4*6

dA/dt = 36+ 24 cm²/s

dA/dt = 60 cm²/s

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

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step-by-step explanation: