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. Read more

Steps to Solve

1. 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.

2. 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 ($dz$): $4 - 3 = 1$

3. 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 $z$: $$ \frac{\partial T}{\partial z} = 190z $$

4. 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 $\frac{\partial T}{\partial z}$: $$ \frac{\partial T}{\partial z}(1, 50, 3) = 190(3) = 570 $$

5. 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:

   $$ dT = 70.18 \cdot 1 + 1.4 \cdot 5 + 570 \cdot 1 $$

   $$ dT = 70.18 + 7 + 570 = 647.18 $$

6. 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:

   $$ y - x^2 - 1 = 0 \implies y = x^2 + 1 $$