The temperature function for a bird in flight is given by T(x,y,z) = 0.09x^2 + 1.4xy + 95z^2. Use differential dT to approximate change in temperature when head wind x increases fr... The temperature function for a bird in flight is given by T(x,y,z) = 0.09x^2 + 1.4xy + 95z^2. Use differential dT to approximate change in temperature when head wind x increases from 1 m/s to 2 m/s, bird heart rate y increases from 50 beats per minute to 55 beats per minute and flapping rate z increases from 3 flaps per second to 4 flaps per second. Consider the function f(x,y) = e^(y-x^2-1). Find the level curve of f that passes through the point (2, 5). Also sketch this curve.
Understand the Problem
The question consists of two parts: (a) involves using the temperature function to approximate changes based on specified increments in three variables, while (b) requires finding the level curve of a given function at a certain point and sketching it.
Answer
The approximate change in temperature is $647.18$ and the level curve is given by $y = x^2 + 1$.
Answer for screen readers
The approximate change in temperature, $dT$, is $647.18$.
The level curve of the function at the point $(2, 5)$ is given by the equation (y = x^2 + 1).
Steps to Solve
-
Identify the Variables and the Temperature Function
The temperature function is given by:
$$ T(x, y, z) = 0.09x^2 + 1.4xy + 95z^2 $$
Here,
- $x$ represents the head wind in m/s,
- $y$ represents the heart rate in beats per minute,
- $z$ represents the flapping rate in flaps per second.
-
Define the Variable Changes
From the problem, we need the changes in the variables:
- Change in head wind ($dx$): $2 - 1 = 1$
- Change in heart rate ($dy$): $55 - 50 = 5$
- Change in flapping rate ($dz$): $4 - 3 = 1$
-
Calculate the Partial Derivatives
Next, we need the partial derivatives of $T$:
-
For $x$: $$ \frac{\partial T}{\partial x} = 0.18x + 1.4y $$
-
For $y$: $$ \frac{\partial T}{\partial y} = 1.4x $$
-
For $z$: $$ \frac{\partial T}{\partial z} = 190z $$
-
-
Evaluate the Partial Derivatives at the Given Point (1, 50, 3)
Substituting $x = 1$, $y = 50$, and $z = 3$:
-
Compute $\frac{\partial T}{\partial x}$: $$ \frac{\partial T}{\partial x}(1, 50, 3) = 0.18(1) + 1.4(50) = 0.18 + 70 = 70.18 $$
-
Compute $\frac{\partial T}{\partial y}$: $$ \frac{\partial T}{\partial y}(1, 50, 3) = 1.4(1) = 1.4 $$
-
Compute $\frac{\partial T}{\partial z}$: $$ \frac{\partial T}{\partial z}(1, 50, 3) = 190(3) = 570 $$
-
-
Calculate the Total Differential (dT)
The total differential $dT$ is given by:
$$ dT = \frac{\partial T}{\partial x} dx + \frac{\partial T}{\partial y} dy + \frac{\partial T}{\partial z} dz $$
Substitute the values we computed:
$$ dT = 70.18 \cdot 1 + 1.4 \cdot 5 + 570 \cdot 1 $$
$$ dT = 70.18 + 7 + 570 = 647.18 $$
-
Level Curve for part (b)
To find the level curve of the function ( f(x, y) = e^{y - x^2 - 1} ) through the point ( (2, 5) ):
- Evaluate ( f(2, 5) ):
$$ f(2, 5) = e^{5 - 2^2 - 1} = e^{5 - 4 - 1} = e^0 = 1 $$
- The level curve equation is:
$$ e^{y - x^2 - 1} = 1 $$
This simplifies to:
$$ y - x^2 - 1 = 0 \implies y = x^2 + 1 $$
The approximate change in temperature, $dT$, is $647.18$.
The level curve of the function at the point $(2, 5)$ is given by the equation (y = x^2 + 1).
More Information
The temperature change calculated shows how sensitive the temperature function is to changes in the head wind, heart rate, and flapping rate. The level curve (y = x^2 + 1) indicates that for any given $x$ value, there’s a corresponding $y$ value that keeps the function output constant.
Tips
- Forgetting to evaluate the partial derivatives at the specified point can lead to incorrect total differentials.
- Miscalculating the changes in each variable can alter the final output.
- Failing to correctly set up the equation for the level curve based on the given function can lead to confusion.
AI-generated content may contain errors. Please verify critical information