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1/f = 1/p + 1/q
0 = -1/p^2 dp/dq - 1/q^2
dp/dq = -p^2/q^2
1f=1p+1q
What is the rate of change of p with respect to q if q=4 and f=2? (Make sure you have the correct sign for the rate.)
0 = -1/p^2 dp/dq - 1/q^2
dp/dq = -p^2/q^2
Given: 1/f = 1/p + 1/q
Differentiating both sides with respect to q:
d(1/f)/dq = d(1/p)/dq + d(1/q)/dq
Since f is a constant, the derivative of 1/f with respect to q is zero.
0 = d(1/p)/dq + d(1/q)/dq
Now, substituting the given values of q=4 and f=2 into the equation, we can solve for the rate of change.
0 = d(1/p)/dq + d(1/4)/dq
Since 1/4 is a constant, the derivative of 1/4 with respect to q is also zero.
0 = d(1/p)/dq + 0
Thus, the rate of change of p with respect to q when q=4 and f=2 is zero.