Given that x=sinht and y=-cosht.
find: dx/dy
dy/dx = ([d/dy]y(t)) / ([d/dt]x(t))
[d/dt]x(t) = [d/dt]sinh(t)=cosh(t)
[d/dt]y(t) = [d/dt]-cosh(t)=-sinh(t)
dy/dx = -sinh(t)/cosh(t)
To find dx/dy given that x = sinh(t) and y = -cosh(t), we can use the chain rule.
Step 1: Differentiate x = sinh(t) with respect to t.
Using the derivative formula for sinh(t), we get:
dx/dt = cosh(t).
Step 2: Differentiate y = -cosh(t) with respect to t.
Using the derivative formula for cosh(t), we get:
dy/dt = -sinh(t).
Step 3: Find dx/dy.
To find dx/dy, we need to find dt/dy first.
Using the chain rule, we have:
dt/dy = 1 / (dy/dt).
Substituting dy/dt = -sinh(t), we get:
dt/dy = 1 / (-sinh(t)).
Since x = sinh(t) and y = -cosh(t), we can substitute them back in:
dt/dy = 1 / (-sinh(arcsinh(x))) = -1 / x.
Finally, to find dx/dy, we use the chain rule again:
dx/dy = (dx/dt) * (dt/dy) = cosh(t) * (-1 / x).
Substituting cosh(t) = sqrt(x^2 + 1) and x = sinh(t), we get:
dx/dy = (-sqrt(x^2 + 1)) / x.
Therefore, dx/dy = (-sqrt(x^2 + 1)) / x.
To find dx/dy, we need to express x in terms of y, and then differentiate it with respect to y.
Given x = sinh(t) and y = -cosh(t), we can rearrange the equations to solve for t in terms of x and y.
From the definition of sinh, we have sinh(t) = x. Rearranging this equation, we get t = sinh^(-1)(x), where sinh^(-1) is the inverse hyperbolic sine function.
Similarly, from the definition of cosh, we have cosh(t) = -y. Rearranging this equation, we get t = cosh^(-1)(-y), where cosh^(-1) is the inverse hyperbolic cosine function.
Since t represents the same value in both equations, we have sinh^(-1)(x) = cosh^(-1)(-y).
Differentiating both sides of the equation with respect to y, we get:
d/dy[sinh^(-1)(x)] = d/dy[cosh^(-1)(-y)]
To differentiate the left-hand side, we need to apply the chain rule. Let's denote sinh^(-1)(x) as u and differentiate it with respect to y:
du/dy = d/dx[sinh^(-1)(x)] * dx/dy
Similarly, for the right-hand side, let's denote cosh^(-1)(-y) as v and differentiate it with respect to y:
dv/dy = d/dx[cosh^(-1)(-y)] * dx/dy
Since u and v are equal (according to the equation we derived earlier), we can equate du/dy and dv/dy:
du/dy = dv/dy
Now we can rearrange the equation to solve for dx/dy:
dx/dy = du/dy / (d/dx[sinh^(-1)(x)])
Differentiating sinh^(-1)(x) with respect to x gives:
d/dx[sinh^(-1)(x)] = 1 / sqrt(1+x^2)
Plugging this back into the equation, we have:
dx/dy = du/dy / (1 / sqrt(1+x^2))
Note: This step assumes that x is not a constant and can vary with respect to y. If x is a constant, then dx/dy would be zero.
So, the expression for dx/dy is:
dx/dy = du/dy * sqrt(1+x^2)
Please note that the derivatives of sinh^(-1)(x) and cosh^(-1)(-y) can be computed using standard rules for differentiating inverse hyperbolic functions.