Use the Chain Rule to find dw/dt.
w = ln (x^2 + y^2 + z^2)^.5
, x = 9 sin t, y = 6 cos t, z = 7 tan t
dw/dt = ∂w/∂x dx/dt + ∂w/∂y dy/dt + ∂w/∂z dz/dt
Using ' to mean d/dt,
w = 1/2 ln(x^2+y^2+z^2)
dw/dt = 1/[2(x^2+y^2+z^2)] * (2xx'+2yy'+2zz')
= ((9sin t)(9cos t)+(6cos t)(-6sin t)+(7tan t)(7sec^2 t))/(81sin^2 t + 36cos^2 t + 7tan^2 t)
= (45sint*cost + 49tant*sec^2 t)/(81sin^2 t + 36cos^2 t + 49tan^2 t)
Don't think you can simplify that much.
Well, well, well, it seems we have a little calculus problem here! Don't worry, I've got my clown shoes on and I'm ready to help you out. Let's grab our chains and get started!
First, let's replace the values of x, y, and z into the expression for w:
w = ln((9 sin t)^2 + (6 cos t)^2 + (7 tan t)^2)^.5
Now, let's simplify that expression inside the logarithm by expanding and simplifying:
w = ln(81 sin^2 t + 36 cos^2 t + 49 tan^2 t)^.5
Next, we're going to use the chain rule to find dw/dt. The chain rule states that if we have a function f(g(t)), then its derivative is given by f'(g(t)) * g'(t).
In this case, our function is w, and our g(t) consists of x, y, and z. So, let's go step by step:
First, we find the derivative of w with respect to g(t):
dw/dg(t) = 0.5 * (81 sin^2 t + 36 cos^2 t + 49 tan^2 t)^(-0.5) * (d/dt(81 sin^2 t + 36 cos^2 t + 49 tan^2 t))
Next, we find the derivative of our g(t) with respect to t. Let's calculate each term's derivative separately:
d(x)/dt = 9 cos t
d(y)/dt = -6 sin t
d(z)/dt = 7 sec^2 t
Now, let's substitute those derivatives into our expression for dw/dg(t):
dw/dg(t) = 0.5 * (81 sin^2 t + 36 cos^2 t + 49 tan^2 t)^(-0.5) * (d/dt(81 sin^2 t + 36 cos^2 t + 49 tan^2 t))
= 0.5 * (81 sin^2 t + 36 cos^2 t + 49 tan^2 t)^(-0.5) * (d(81 sin^2 t)/dt + d(36 cos^2 t)/dt + d(49 tan^2 t)/dt)
= 0.5 * (81 sin^2 t + 36 cos^2 t + 49 tan^2 t)^(-0.5) * (162 sin t cos t - 72 sin t cos t + 98 sec^2 t tan t)
Finally, we have dw/dg(t), but we want dw/dt. So, let's substitute our expressions for dx/dt, dy/dt, and dz/dt:
dw/dt = dw/dg(t) * dx/dt + dw/dg(t) * dy/dt + dw/dg(t) * dz/dt
= [(162 sin t cos t - 72 sin t cos t + 98 sec^2 t tan t) * 9 cos t + (162 sin t cos t - 72 sin t cos t + 98 sec^2 t tan t) * (-6 sin t) + (162 sin t cos t - 72 sin t cos t + 98 sec^2 t tan t) * 7 sec^2 t]
And voila! That's the expression for dw/dt using the Chain Rule. Feel free to simplify it further if you like, but remember that clown math is all about having fun with numbers!
To find dw/dt using the Chain Rule, we'll break down the problem into smaller steps.
Step 1: Find the partial derivatives of w with respect to x, y, and z.
w = ln (x^2 + y^2 + z^2)^.5
Let's find the partial derivative of w with respect to x, keeping y and z constant:
∂w/∂x = (∂w/∂(x^2 + y^2 + z^2)) * (∂(x^2 + y^2 + z^2)/∂x)
∂w/∂x = (1/(x^2 + y^2 + z^2)^.5) * (∂(x^2 + y^2 + z^2)/∂x)
Since x = 9 sin t, y = 6 cos t, and z = 7 tan t, we can substitute these values into the equation:
∂w/∂x = (1/(9^2 sin^2 t + 6^2 cos^2 t + 7^2 tan^2 t)^.5) * (∂(9^2 sin^2 t + 6^2 cos^2 t + 7^2 tan^2 t)/∂x)
Step 2: Simplify the expression and find the derivative.
∂w/∂x = (1/(81 sin^2 t + 36 cos^2 t + 49 tan^2 t)^.5) * (∂(81 sin^2 t + 36 cos^2 t + 49 tan^2 t)/∂x)
Since we are differentiating with respect to x, we only need to find the derivative of 81 sin^2 t:
∂w/∂x = (1/(81 sin^2 t + 36 cos^2 t + 49 tan^2 t)^.5) * (2(9 sin t))
Simplifying further:
∂w/∂x = (18 sin t)/(81 sin^2 t + 36 cos^2 t + 49 tan^2 t)^.5
Now, you can repeat the above steps to find ∂w/∂y and ∂w/∂z. Once you have found all the partial derivatives, you can use the Chain Rule to find dw/dt.
To find dw/dt using the Chain Rule, we need to differentiate w with respect to each variable (x, y, z) and then multiply them by the derivatives of x, y, and z with respect to t.
First, we differentiate w with respect to x. The derivative of ln(u) is 1/u multiplied by the derivative of u with respect to x. In this case, u = (x^2 + y^2 + z^2)^0.5. So, using the chain rule, dw/dx = (1/u) * du/dx.
Next, we differentiate w with respect to y. Again, using the chain rule, dw/dy = (1/u) * du/dy.
Finally, we differentiate w with respect to z. Once again, using the chain rule, dw/dz = (1/u) * du/dz.
Now, let's find the derivatives of x, y, and z with respect to t:
dx/dt = d(9 sin t)/dt = 9 cos t
dy/dt = d(6 cos t)/dt = -6 sin t
dz/dt = d(7 tan t)/dt = 7 sec^2 t
Substituting these derivatives into the expressions we derived earlier, we get:
dw/dt = (1/u) * (du/dx * dx/dt + du/dy * dy/dt + du/dz * dz/dt)
Substituting the values for x, y, and z, we have:
u = (x^2 + y^2 + z^2)^0.5 = ((9 sin t)^2 + (6 cos t)^2 + (7 tan t)^2)^0.5
du/dx = d/dx(sqrt(x^2 + y^2 + z^2)) = (1/2)(x^2 + y^2 + z^2)^(-0.5)(2x)
du/dy = d/dy(sqrt(x^2 + y^2 + z^2)) = (1/2)(x^2 + y^2 + z^2)^(-0.5)(2y)
du/dz = d/dz(sqrt(x^2 + y^2 + z^2)) = (1/2)(x^2 + y^2 + z^2)^(-0.5)(2z)
Now, we can substitute all the values into our expression for dw/dt:
dw/dt = (1/u) * [(du/dx * dx/dt) + (du/dy * dy/dt) + (du/dz * dz/dt)]
= (1/u) * [(1/2)(x^2 + y^2 + z^2)^(-0.5)(2x) * 9 cos t + (1/2)(x^2 + y^2 + z^2)^(-0.5)(2y) * -6 sin t + (1/2)(x^2 + y^2 + z^2)^(-0.5)(2z) * 7 sec^2 t]
= [(x cos t - y sin t + z sec^2 t) / (2u(x^2 + y^2 + z^2)^(0.5))]
= [(9sin(t)cos(t) - 6cos(t)sin(t) + 7tan(t)sec^2(t)) / (2(x^2 + y^2 + z^2)^(0.5))]
Therefore, dw/dt = [(9sin(t)cos(t) - 6cos(t)sin(t) + 7tan(t)sec^2(t)) / (2(x^2 + y^2 + z^2)^(0.5))]