- #1
Alex Bard
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Hi, I have a test prep question regarding Chain Rule, please see the problem and my attempt below. I believe part A is okay but part B, I'm just confused, seems like there is a part missing from the question, or at least how I'm use to doing it.
[itex]A. Let f(x, y) = cos(xy) + ycos(x), where x = u^2 + v and y = u - v^2. Find \frac{∂z}{∂v} when u = 1, v = -1.[/itex]
B. Let f(x,y) = ln(x - 3y). Find f(6.9, 2.06)
Chain Rule is stated by [itex] \frac{∂z}{∂x} * \frac{∂x}{∂v} + \frac{∂z}{∂y} * \frac{∂y}{∂v}[/itex]
Part A.
[itex]frac{∂z}{∂v}[/itex] = cos(xy) + y cos(x)
First take the partial in respect to x and multiply by partial to respect to v
= (y)[-sin(xy) - y sin(x)] (2u + v)
Take partial in respect to y and multiply by partial in respect to v
= (x)[-sin(xy) + cos(x)] (u - 2v)
Add both of them to complete:
= (y)[-sin(xy) - y sin(x)] (2u + v) + (x)[-sin(xy) + cos(x)] (u - 2v)
Now, I substitute the values of x & y with their respective counterparts:
= (u - v^2)[-sin((u^2+v)(u-v^2)) - (u-v^2)(sin(u^2+v))] + (u^2+v)[-sin((u^2+v)(u-v^2)) + cos(u^2+v)](u-2v)
Now I will plug in the respective values for u & v with their respective counterparts:
= (1 - (-1)^2)[-sin((1^2+(-1))(1-(-1)^2)) - (1-(-1)^2)(sin(1^2+(-1)))] + (1^2+(-1))[-sin((1^2+(-1))(1-(-1)^2)) + cos(1^2+(-1))](1-2(-1))
Now to simplify and Solve:
Now I will plug in the respective values for u & v with their respective counterparts:
= (0)[-sin((0)(0)) - (0)(sin(0))] + (0)[-sin((0)(0)) + cos(0)](3)
= 0 + 0 (3)
so the answer I get is 0.
Does that seem right?
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Part B.
Let f(x, y) = ln(x - 3y). Find (6.9, 2.06)
I believe this problem is missing a point but considering this problem as is, I got:
Find f(6.9, 2.06) = ln(x - 3y)
→ (6.9 - 6.18) = 0.72 → ln(0.72) ≈ -0.33
Now zx = [itex]\frac{1}{x - 3y}[/itex]
Now zy = [itex]\frac{-3}{x - 3y}[/itex]
so my final answer is
[itex]\frac{1}{.72} , \frac{-3}{.72}[/itex]
Thank you for your help.
Homework Statement
[itex]A. Let f(x, y) = cos(xy) + ycos(x), where x = u^2 + v and y = u - v^2. Find \frac{∂z}{∂v} when u = 1, v = -1.[/itex]
B. Let f(x,y) = ln(x - 3y). Find f(6.9, 2.06)
Homework Equations
Chain Rule is stated by [itex] \frac{∂z}{∂x} * \frac{∂x}{∂v} + \frac{∂z}{∂y} * \frac{∂y}{∂v}[/itex]
The Attempt at a Solution
Part A.
[itex]frac{∂z}{∂v}[/itex] = cos(xy) + y cos(x)
First take the partial in respect to x and multiply by partial to respect to v
= (y)[-sin(xy) - y sin(x)] (2u + v)
Take partial in respect to y and multiply by partial in respect to v
= (x)[-sin(xy) + cos(x)] (u - 2v)
Add both of them to complete:
= (y)[-sin(xy) - y sin(x)] (2u + v) + (x)[-sin(xy) + cos(x)] (u - 2v)
Now, I substitute the values of x & y with their respective counterparts:
= (u - v^2)[-sin((u^2+v)(u-v^2)) - (u-v^2)(sin(u^2+v))] + (u^2+v)[-sin((u^2+v)(u-v^2)) + cos(u^2+v)](u-2v)
Now I will plug in the respective values for u & v with their respective counterparts:
= (1 - (-1)^2)[-sin((1^2+(-1))(1-(-1)^2)) - (1-(-1)^2)(sin(1^2+(-1)))] + (1^2+(-1))[-sin((1^2+(-1))(1-(-1)^2)) + cos(1^2+(-1))](1-2(-1))
Now to simplify and Solve:
Now I will plug in the respective values for u & v with their respective counterparts:
= (0)[-sin((0)(0)) - (0)(sin(0))] + (0)[-sin((0)(0)) + cos(0)](3)
= 0 + 0 (3)
so the answer I get is 0.
Does that seem right?
-----------------------------------------------------------------------------------------------------------
Part B.
Let f(x, y) = ln(x - 3y). Find (6.9, 2.06)
I believe this problem is missing a point but considering this problem as is, I got:
Find f(6.9, 2.06) = ln(x - 3y)
→ (6.9 - 6.18) = 0.72 → ln(0.72) ≈ -0.33
Now zx = [itex]\frac{1}{x - 3y}[/itex]
Now zy = [itex]\frac{-3}{x - 3y}[/itex]
so my final answer is
[itex]\frac{1}{.72} , \frac{-3}{.72}[/itex]
Thank you for your help.
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