A balloon is rising vertically above a level straight Rd. at a constant rate of four feet per second. So our change in height in terms of time is going to be 4 feet per second or DYDT equal 4 feet per second. Just when the balloon is 60 feet above the ground, a bicycle moving at a constant rate of 18 feet per second passes under it. So our DX DT is going to be 18 feet per second and our Y is currently 60 feet. How fast is the distance as of T between the bicycle and the balloon increasing 3 seconds later? So at time zero, the bicycle is directly underneath the balloon and the balloon is 60 feet above it. So 3 seconds later the bicycle has gone 3 seconds at 18 feet per second or 54 feet. The balloon has gone its original 60 + 3 seconds at 4 feet per second. So the balloons height is now 72 feet. We're going to find the distance between the balloon and the bicycle by using Pythagorean theorem. So our S at three is going to equal sqrt 54 ^2 + 72 ^2. Computing that out, we get 90. Now we know that it's a right triangle, so X ^2 + y ^2 = s ^2. Taking the derivative, we have D2XD XD t + 2 YDYDT equaling 2S DSDT. Dividing everything through by A2, we get XDXD t + y DYDT equaling SDSDT. Plugging in our knowns, our X is 54, our DXDT is 18, our Y is 72, our DYDT is 4, our S is 90. We're going to find our DSDT, so we just multiply, add and then divide and we get 14 feet per second.