Let l be the length of a diagonal of a rectangle whose sides have lengths x and y, and assume that x and y vary with time.

Question

if x increases at a constant rate of 1/6ft/s and y decreases at a constant rate of 1/8 ft/s, how fast is the size of the diagonal changing when x = 4 ft and y = 4 ft?

 (Round your answer to four decimal places. Be sure to choose the correct units.)

Answer 1

Hi Natalie!

When the diagonal is first drawn, we will notice that it forms a right triangle with lengths x and y of the rectangle. The relationship between the three sides in a right triangle can be expressed using the Pythagorean Theorem. They say the diagonal has length "L", so we will use this variable as the hypotenuse.

x² + y² = L²

Now, let's differentiate this equation with respect to time, so that we have it set up to find out how fast the length is changing:

x² (d/dt) + y² (d/dt) = L² (d/dt)

Using the chain rule to complete the differentiation, the derivative equation is:

2x (dx/dt) + 2y (dy/dt) = 2L (dL/dt)

They tell us that x is increasing at a constant rate of 1/6 ft./sec. and that y is decreasing (negative) at a constant rate of 1/8 ft./sec. Therefore:

dx/dt = 1/6 ft./sec. and

dy/dt = -1/8 ft./sec.

Now, let's find out what the length of the diagonal is (L) when x = 4 and y = 4 before we solve for dL/dt. We can use the Pythagorean Theorem to figure this out.

x² + y² = L²

4² + 4² = L² ----> 16 + 16 = L²

L² = 32.

L = sqrt(32)

Therefore, the length of the diagonal is the sqrt(32), which simplified is 4*sqrt(2)

Now that we know the values of all of the unknowns except for dL/dt, the rate in which the length of the diagonal is changing, from our derivative equation, we can substitute and solve for dL/dt when x = 4 and y = 4. Using our derivative equation that we got up above:

2x (dx/dt) + 2y (dy/dt) = 2L (dL/dt)

2(4)(1/6) + 2(4)(-1/8) = 2(4*sqrt(2)) dL/dt

(4/3) + (-1) = 8*sqrt(2) dL/dt

1/3 = 8*sqrt(2) dL/dt

1/(24*sqrt(2)) = dL/dt

dL/dt = 0.0295 ft./sec (rounded to 4 decimal places)

In conclusion, the rate in which the diagonal's length (dL/dt) is changing (specifically increasing in this case as it is a positive value) is 0.0295 ft./sec.

Let me know if you have any questions and if that helps.

Thanks!

Answer 2