If f is the focal length of a convex lens and an object is placed at a distance p from the lens, then its image will be at a distance q from the lens, where f, p, and q are related by the lens equation
1/f = 1/p + 1/q
What is the rate of change of p with respect to q if q = 2 and f = 6? (Make sure you have the correct sign for the rate.)
STANDARD form given slope is 2/3 and have a y intercept of -4
Well, looks like we've got ourselves some lens equation fun!
Let's differentiate the lens equation with respect to q and solve for dp/dq:
d(1/f)/dq = d(1/p)/dq + d(1/q)/dq
Since we know that f = 6 and q = 2, we can plug those values in and simplify:
d(1/6)/dq = d(1/p)/dq + d(1/2)/dq
Simplifying further:
-1/6^2 * dp/dq = -1/p^2 * dp/dq + 0
Since we're trying to find dp/dq, let's solve for it:
-1/36 * dp/dq = -1/p^2 * dp/dq
If we bring both terms with dp/dq to the left side, we get:
-1/36 * dp/dq + 1/p^2 * dp/dq = 0
Factor out dp/dq:
(1/p^2 - 1/36) * dp/dq = 0
To find the rate of change of p with respect to q, we can solve for dp/dq:
dp/dq = 0 / (1/p^2 - 1/36)
The denominator is equal to:
1/p^2 - 1/36 = (36 - p^2) / (36p^2)
So, we can rewrite dp/dq as:
dp/dq = 0 / ((36 - p^2) / (36p^2))
Since anything divided by 0 is undefined, the rate of change of p with respect to q is undefined when q = 2 and f = 6.
Well, isn't that funny? It seems like the rate of change of p with respect to q is not defined in this case. Keep in mind that this is just a mathematical interpretation, and I apologize if it's not the laugh you were hoping for!
To find the rate of change of p with respect to q, we need to differentiate the lens equation with respect to q:
d(1/f)/dq = d(1/p)/dq + d(1/q)/dq
Since f and p are constants in this problem, their derivatives will be zero. Therefore, the equation simplifies to:
0 = 0 + d(1/q)/dq
To find d(1/q)/dq, we can differentiate 1/q with respect to q:
d(1/q)/dq = -1/q^2
Now we can substitute the values of q = 2 and f = 6 into the expression:
d(1/q)/dq = -1/(2^2) = -1/4
Therefore, the rate of change of p with respect to q is -1/4.
To find the rate of change of p with respect to q, we need to take the derivative of the lens equation with respect to q. Let's differentiate both sides of the equation:
1/f = 1/p + 1/q
Differentiating both sides with respect to q:
0 = -1/p^2 - 1/q^2 * dq/dp
Now, rearrange the equation to solve for dq/dp:
dq/dp = -q^2/(p^2 * (1/f))
Substitute the given values: f = 6 and q = 2:
dq/dp = -2^2 / (p^2 * (1/6))
Simplifying the expression:
dq/dp = -4 / (p^2 / 6)
Now, if we want to find the rate of change of p with respect to q when q = 2, we substitute q = 2 into the equation and solve for dp/dq:
dq/dp = -4 / (p^2 / 6)
2 = -4 / (p^2 / 6)
Cross-multiply and simplify the equation:
2 * (p^2 / 6) = -4
(p^2 / 3) = -2
Multiply both sides of the equation by 3 to solve for p^2:
p^2 = -6
Since p^2 cannot be negative, there is no real solution to this equation. Therefore, the rate of change of p with respect to q is undefined when q = 2 and f = 6.
1/6 = 1/p + 1/2
1/p = 1/6 - 3/6 = - 2/6 = -1/3
so
p = -3
now the calculus, assume f is a constant of the lens so df/d anything = 0.
0 = (-dp/dq)/p^2 - (-1)/q^2
dp/dq = p^2/q^2 = 9/2 = 4.5