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Related Rates
- Thread starter sar_a215
- Start date
1. The length of a rectangle is increasing at a rate of 8cm/s and its width is decreasing at a rate of 3cm/s. At what rate is the area of the rectangle changing when the width is 10cm and the length is 20cm?
Always start with a sketch. Let’s draw a rectangle on an xy-graph in the first quadrant. Place the lower, left-hand corner of the rectangle on the origin.
Let the length be the x-side of the rectangle (x = 20 cm) and the width be the y-side (y = 10 cm). Label this on your drawing. Therefore,
A = xy = (20 cm)(10 cm) = 200 cm^2
At the lower, right-hand corner of the rectangle, draw a short arrow pointing to the right (the positive x direction). Underneath the arrow write dx/dt = 8 cm/s.
At the upper, left-hand corner of the rectangle, draw a short arrow pointing down (the negative y direction). Next to the arrow, write dy/dt = -3 cm/s.
Please take a moment to think about these last two steps. Do you understand why dx/dt = 8 cm/s and dy/dt = -3 cm/s?
In related rates problems, we are always differentiating with respect to time, t, -- not with respect to x, as you are probably used to doing. Pursuant to this, let’s go back to our area equation and differentiate it with respect to time:
A = xy
dA/dt = y(dx/dt) + x(dy/dt)
The problem asked us to find “At what rate is the area of the rectangle changing”. That means find dA/dt. So, now we just plug everything in:
dA/dt = y(dx/dt) + x(dy/dt) = (10 cm)(8 cm/s) + (20 cm)(-3 cm/s) = 20 cm^2/s
The sign of the answer is positive, indicating the area is growing. Note that our final units are in terms of area (cm^2) per unit of time (s). Make sense?
A = xy
dA/dt = y(dx/dt) + x(dy/dt)
so this question doesn't have use of the product rule at all?
Look at the above. This IS the product rule.