One diagonal of a rhombus is decreasing at a rate of 7 centimeters per minute and the other diagonal of the rhombus is increasing at a rate of 10 centimeters per minute.At a certain instant, the decreasing diagonal is 4 centimeters and the increasing diagonal is 6 centimeters.What is the rate of change of the area of the rhombus at that instant (in square centimeters per minute)?Choose 1 answer:(A) −1(B) 16(C) −16(D) 1The area of a rhombus with diagonals d1 and d2 is 2d1d2.
Q. One diagonal of a rhombus is decreasing at a rate of 7 centimeters per minute and the other diagonal of the rhombus is increasing at a rate of 10 centimeters per minute.At a certain instant, the decreasing diagonal is 4 centimeters and the increasing diagonal is 6 centimeters.What is the rate of change of the area of the rhombus at that instant (in square centimeters per minute)?Choose 1 answer:(A) −1(B) 16(C) −16(D) 1The area of a rhombus with diagonals d1 and d2 is 2d1d2.
Rhombus Area Formula: The formula for the area of a rhombus is (d1×d2)/2, where d1 and d2 are the lengths of the diagonals.
Diagonal Lengths Given: Let's denote the decreasing diagonal as d1 and the increasing diagonal as d2. At the instant we are considering, d1=4cm and d2=6cm.
Rates of Change: The rate of change of d1 is −7cm/min (since it's decreasing) and the rate of change of d2 is 10cm/min (since it's increasing).
Rate of Change Formula: To find the rate of change of the area, we'll use the product rule for differentiation, which in this context is dtd(Area)=2(d1⋅dtd(d2)+d2⋅dtd(d1)).
Calculation: Plugging in the values, we get dtd(Area)=2(4cm×10cm/min+6cm×−7cm/min).
Final Result: Now, let's do the calculation: dtd(Area)=2(40−42)=2−2=−1 cm2/min.
More problems from Area of quadrilaterals and triangles: word problems