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