Question

if you have a box and the lenghth is 12cm and the width is 5cm and the width is increasing at 2 cm/sec and the length is decreasing at 2 cm/sec what is the DA/dt for the area perimeter and diagonal

Answer 1

A=l*b
dA/dt = d(l*b)/dt=dl/dt *b + l*db/dt= -2b+2l

Answer 2

i honestly have no idea which variable is what

Answer 3

l is length and b is width

Answer 4

okay so area is 14cm/sec and perimiter is 0 but the diagonal?

Answer 5

diagonal^2 = l^2 +b^2
x^2=l^2+b^2
2x dx/dt = 2l dl/dt + 2b db/dt

Answer 6

where x= diagonal

Answer 7

Does it help?

Answer 8

okay so pythagerous

Answer 9

pythagorus*

Answer 10

can the answer be negative? amd do you know the answer?

Answer 11

yes the answer can be negative

Answer 12

x^2=l^2+b^2
x^2=12^2 +5^2
x=13
Now,
2x dx/dt = 2l dl/dt + 2b db/dt
2*13 dx/dt = 2*12*(-2) +2*5*2

Answer 13

U need to find dx/dt which looks like is negative