Question

One base of a trapezoid is decreasing at a rate of 888 kilometers per second and the height of the trapezoid is increasing at a rate of 555 kilometers per second. The other base of the trapezoid is fixed at 444 kilometers. At a certain instant, the decreasing base is 121212 kilometers and the height is 222 kilometers. What is the rate of change of the area of the trapezoid at that instant (in square kilometers per second)

Answer 1

The rate of change of the area of the trapezoid at that instant is 5 square kilometers per second and this can be determined by differentiating the area of the trapezoid with respect to time 't'.

Given :

• One base of a trapezoid is decreasing at a rate of 8 kilometers per second and the height of the trapezoid is increasing at a rate of 5 kilometers per second.
• The other base of the trapezoid is fixed at 4 kilometers.
• At a certain instant, the decreasing base is 12 kilometers and the height is 2 kilometers.

First, determine the area of a trapezoid, that is:

$\rm A = \dfrac{b_1+b_2}{2}\times h$

$\rm \dfrac{2A}{h}=b_1+b_2$

$\rm b_2 = \dfrac{2A}{h}-b_1$

Now, differentiate the above expression with respect to 't'.

$\rm \dfrac{db_2}{dt} = -\dfrac{2A}{h^2}\dfrac{dh}{dt}-\dfrac{db_1}{dt}$

Now. substitute the values of the known terms in the above formula.

$\rm \dfrac{db_2}{dt} = -\dfrac{2\times (16)}{2^2}\times (5)-(-8)$

$\rm \dfrac{db_2}{dt} = -32\;km/sec$

So, the area is decreasing at the rate of 5.

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https://brainly.com/question/1083374