Question 1172159: Solve the following related rates problem.Show your complete solution
1. A spherical snowball is melting in such a way that its surface area decreases at the rate of 1 in2/min.
(a) How fast is its radius shrinking when it is 3 in?
(b) How fast is its volume shrinking when its radius is 3 in?
2. Two cars begin a trip from the same point P. If car A travels north at the rate of 30 mi/h and car B travels west at the rate of 40 mi/h, how fast is the distance between them changing 2 hours later?
3. Ship A is 70 km west of ship B and is sailing south at the rate of25 km/h. Ship B is sailing north at the rate of 45 km/h. How fast is the distance between the two ships changing 2 hours later?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! A=4πr^2 in^2
dA/dt=8πr dr/dt units in^2/min
-1/8πr=dr/dt
but r=3 in
so -1/24π is the rate of change in inches of the radius when it is 3 inches, or -0.131 in/min
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Volume is (4/3)πr^3
dV/dt=(4πr^2)dr/dt=4π*9*(-1/24π)
=-1.5 in^3/min
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Draw this triangle, with the x-axis length 80 mi and rate of change 40 mi/hr and the y-axis 60 mi and 30 mi/hr.
z^2=x^2+y^2
2z dz/dt=2x dx/dt+ 2y dy/dt
or dz/dt=(x dx/dt+y dy/dt)/z, dividing first by 2 and then by z
=80mi*40 mi/hr+60 * 30 mi/hr divided by 100 mi, since this is a 6-8-10 type triangle
The units will be mi/hr, which is what we want.
this is 5000 mi^2/hr/100 mi
or 50 mi/hr
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This is the same sort of issue. Draw it out, and the hypotenuse is z. The y-axis is 90+50 or 140 km N-S distance. The x-axis is 70 km E-W distance.
so dz/dt=(x dx/dt +y dy /dt)/z, and z is 156.52 km by the Pythagorean Theorem
Here, however, dx/dt is 0. They are still 70 km apart in the E-W plane so the triangle is keeping one leg constant.
dz/dt=140*70/156.52=62.61 km/h opening the distance
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500 feet of fence
x parallel to lake
(500-x) for the other parts, and there are 3, so the rectangle is x*(500-x)/3 or x*500/3 -x^2/3
the maximum area is where x=-b/2a or (-500/3)/(-2/3)=250 ft.
so x is 250 feet, and the three equal sides are 250/3 or 83.3 feet each.
The area is 20,833.33 sq ft (not rounded)
2.An open box is to be constructed from a 12- × 12-inch piece of cardboard by cutting away squares of equal size from the four corners and folding up the sides. Determine the size of the cutout that maximizes the volume of the box.
x=side of cut out square
each square cut out is x inches on a side. Folded up, that is the height.
the length and width are the same (it is a square) and each is 12-2x inches
so the volume is (12-2x)(12-2x)*x
=(144-48x+4x^2)*x
=144x-48x^2+4x^3
can graph that or take the derivative and set equal to 0 to see what the maximum area is.
derivative is 12x^2-96x+144=0
12(x^2-8x+12)=0
(x-6)(x-2)=0
the squares cut out are either 6 inches on a side (which is impossible since the side is only 12 inches and one has to cut out 2.)
x=2
volume is 2*8*8=128 in ^3.
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