# 8. Solutions for Variations Problems

1. Since x varies as y we can write x=yk where k is proportionality constant which means \(k=\frac{8}{15}.\) Now when y=10 x=10*8/15 i.e. \(x=\frac{16}{3}.\)

2. Since x varies inversely as y we have \(xy=k\) i.e. k=21 since x=3 and y=7. Clearly, when x=7 y will become 3.

3. Given \(x^2=ky^3\) and when x=3, y=4 which implies k=9/64. So our problem is:

4. Since x varies as y and z jointly we have x=yzk. From given values of x, y and z let us compute k.

Now we need to compute z when x=54 and y=3

5. Given a=ck1 and b=ck2 where k1 and k2 are some constants. \(a\pm b\) implies \(c(k1\pm k2)\) therefore it varies as c. Similarly, \(\sqrt{ab}\) implies \(c\sqrt{k1k2}\) therefore it also varies as c.

6. Since a varies as bc can write a=bck. This means b = (a/ck) = ak1/c where k1=1/k and since k is constant k1 will also be constant. The above we can also write as b=k1/(c/a) so we can say b varies inversely as c/a.

From problem statement

8. Since x varies as y x=yk. Clearly \(x^2+y^2=y^2(k^2+1)\) and \(x^2-y^2=y^2(k^2-1)\) now \(k^2+1\) and \(k^2-1\) are constants since k is a constant. Therefore, \(x^2+y^2\) varies as \(x^2-y^2\).

9. Let y=z1+z2 now given that z1=xk1 and z2=x/k2 where k1 and k2 are some constants.

Given when y=6 the x=4. This implies:

Also, when y=10/3 then x=3. This implies

Now these two equations involving k1 and k2 can be solved to get their values and then the relation between x and y can be established.

Rest of the problems are left as exercise to the reader.