our equation for the velocity of the rock is v = −9t + 18. Next, because the height of the rock is
the derivative of the velocity, we can write = −9t + 18. We again separate the variables by
multiplying both sides by dt: dh = (−9t + 18) dt. We integrate both sides:
∫^ dh = ∫ (−9t + 18) dt
h = − + 18t + C 1
Next, we plug in h = 45 and t = 0 to solve for C 1.
45 = − + 18(0) + C 1 , so C 1 = 45
Therefore, the equation for the height of the rock, h, at time t is h = − + 18t + 45.
Finally, we can solve for the height of the rock at time t = 4: h = − + 18t + 45, so h = −
+ 18(4) + 45 = 45m.
- 8,900 grams (approximately)
We can express this situation with the differential equation = −km, where m is the mass at
time t. We are also given that m = 10,000 when t = 0 and m = 5,000 when t = 5,750. We solve
this differential equation by separation of variables. We want to get all of the m variables on
one side of the equals sign and all of the t variables on the other side. We can do this easily by
dividing both sides by m and multiplying both sides by dt. We get = −k dt. Next, we
integrate both sides.
∫^ = ∫ −k dt
ln m = −kt + C 0
