Find dw/dt when w= ln(sqrt(x ² + y

tarynosanders

tarynosanders

Answered question

2022-07-17

Find dw/dt when w= ln(sqrt(x ² + y ² + z ²)) and x = 2 sin(t) y = 8 cos(t) and z = 6 tan (t)

Answer & Explanation

Nick Camelot

Nick Camelot

Skilled2023-06-05Added 164 answers

To find dwdt when w=ln(x2+y2+z2), and x=2sin(t), y=8cos(t), and z=6tan(t), we need to use the chain rule of differentiation.
Let's start by substituting the given expressions for x, y, and z into the expression for w:
w=ln((2sin(t))2+(8cos(t))2+(6tan(t))2)
Simplifying further:
w=ln(4sin2(t)+64cos2(t)+36tan2(t))
Next, we can simplify the expression inside the square root:
w=ln(4sin2(t)+64cos2(t)+36tan2(t))
Using trigonometric identities, we can rewrite the expression as:
w=ln(4(sin2(t)+16cos2(t))+36tan2(t))
w=ln(4+12cos2(t)+36tan2(t))
w=ln(40+12cos2(t)+36tan2(t))
Now, we can differentiate w with respect to t using the chain rule. The derivative dwdt is given by:
dwdt=ddt(ln(40+12cos2(t)+36tan2(t)))
Applying the chain rule, the derivative becomes:
dwdt=140+12cos2(t)+36tan2(t)·ddt(40+12cos2(t)+36tan2(t))
Now, let's differentiate the expression inside the parentheses:
dwdt=140+12cos2(t)+36tan2(t)·(024cos(t)sin(t)+72tan(t)sec2(t))
Simplifying further:
dwdt=24cos(t)sin(t)+72tan(t)sec2(t)40+12cos2(t)+36tan2(t)
Therefore, dwdt=24cos(t)sin(t)+72tan(t)sec2(t)40+12cos2(t)+36tan2(t).

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