1. A 10-foot plank is leaning against a wall. If at a certain instant the bottom
ID: 3187731 • Letter: 1
Question
1. A 10-foot plank is leaning against a wall. If at a certain instant the bottom of the plank is3 feet from the wall and is being pushed toward the wall at the rate of 1/3 foot/sec, how
fast is the acute angle that the plank makes with the ground increasing?
2. Sand is being dropped in a conical pile at a rate of 15 cubic feet per minute. The height
of the pile is always equal to its diameter. How fast is the height increasing when the pile
is 7 feet high?
3. An airplane at a height of 3 miles is flying horizontally at a velocity of 500 miles per hour
and passes directly over a civil defense observer. How fast is the plane moving away
from the observer when it is 5 miles away from the observer?
4. A spherical balloon is inflated so that its volume is increasing at the rate of 3ftcubed/min.
How fast is the diameter of the balloon increasing when the radius is 1 foot?
Find the differential of each of the following.
y = 3x^4-5/x+sqrtx+ csc 3x
y = 3xcos5x
7. Find the equation of the tangent line to the given curve at the given point.
x*y^3 + 2sec x = 2y - 6x + 8 at (0, -3)
Explanation / Answer
1. Py's Theorem: a^2 + b^2 = c^2 Derive that beast with respect to time: 2a(da/dt) + 2b(db/dt) = 2c(dc/dt) c = 10 feet b = 2 feet find a with Py's theorem: a = sqrt(c^2 - b^2) a = sqrt((10)^2 - (2)^2) a = 9.798 feet So now you know a, b and c. db/dt = -6 in/s (because it's sliding towards the wall) dc/dt = 0 in/s (because the ladder is always the same length) So sub all of that in and solve for da/dt, but convert all of the feet to inches by multiplying by 12. 2a(da/dt) + 2b(db/dt) = 2c(dc/dt) 2(9.798 x 12)(da/dt) + 2(2 x 12)(-6in/s) = 2(10 x 12)(0 in/s) 2(9.798 x 12)(da/dt)= 2(2 x 12)(6in/s) da/dt = [2(2 x 12)(6in/s)] / 2(9.798 x 12) da/dt = 1.22 in/s OK so you know the rate of change of all three sides. You can relate them with trigonometry. The angle that the thing makes with the ground is going to be opposite a (call it x). tan(x) = opp/adj tan(x) = 9.798 / 2 x = (tan^-1)(9.798 / 2) x = 78.46 degrees Now use the cos law: a^2 = b^2 + c^2 - 2bc(cosA) Derive all of that with respect to time: 2a(da/dt) = 2b(db/dt) + 2c(dc/dt) - 2(db/dt)(c)(cosA) - 2(b)(dc/dt)(cosA) -2(b)(c)(-sinA)(dA/dt) That's a mess and a half, but you actually know everything except dA/dt: 2(9.798)(1.22)= 2(2)(-6) + 2(10)(0) - 2(-6)(10)(cos78.46) - 2(2)(0)(cos78.46) - 2(2)(10)(-sin78.46)(dA/dt) [2(9.798)(1.22) - 2(2)(-6) + 2(-6)(10)(cos78.46)] / [- 2(2)(10)(-sin78.46)] = dA/dt 0.61 degrees / second = dA/dt
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