414 CHAPTER 3 Polynomial and Rational Functions CAUTION Note that and in the expression “y varies jointly as x and z” translates as the product y = kxz. The word “and” does not indicate addition here. EXAMPLE 3 Solving a Joint Variation Problem The area of a triangle varies jointly as the lengths of the base and the height. A triangle with base 10 ft and height 4 ft has area 20 ft2. Find the area of a triangle with base 3 ft and height 8 ft. (See Figure 69.) SOLUTION Step 1 Let represent the area, b the base, and h the height of the triangle. Then, for some number k, = kbh. varies jointly as b and h. Step 2 is 20 when b is 10 and h is 4, so substitute and solve for k. 20 = k1102142 Substitute for , b, and h. 1 2 = k Solve for k. Step 3 The relationship among the variables is the familiar formula for the area of a triangle, = 1 2 bh. Step 4 To find when b = 3 ft and h = 8 ft, substitute into the formula. = 1 2 132182 = 12 ft2 EXAMPLE 4 Solving a Combined Variation Problem The number of vibrations per second (the pitch) of a steel guitar string varies directly as the square root of the tension and inversely as the length of the string. If the number of vibrations per second is 50 when the tension is 225 newtons and the length is 0.60 m, find the number of vibrations per second when the tension is 196 newtons and the length is 0.65 m. SOLUTION Step 1 Let n represent the number of vibrations per second, T represent the tension, and L represent the length of the string. Then, from the information in the problem, write the variation equation. n = k2T L n varies directly as the square root of T and inversely as L. Step 2 Substitute the given values for n, T, and L and solve for k. 50 = k2225 0.60 Let n = 50, T = 225, L = 0.60. 30 = k2225 Multiply by 0.60. 30 = 15k 2225 = 15 k = 2 Divide by 15. Interchange sides. S Now Try Exercise 39. 10 ft 4 ft Area = 20 ft2 8 ft 3 ft ! = ? Figure 69
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