Find $$ \frac{\partial w}{\partial s} $$ using the appropriate Chain Rule for $$ w=y^{3}-5 x^{2} y $$ where $$ x=e^{3} $$ and $$ y=e^{t} $$, and evaluate the partial derivative at $$ s=-3 $$ and $$ t=5 $$. Round your answer to two decimal places.

Question

UpStudy AI · Accepted Answer

We start with the function
$$
w = y^3 - 5x^2 y,
$$
and we are given
$$
x = e^{3s}\quad \text{and} \quad y = e^t.
$$
Thus the function $$w$$ becomes a function of $$s$$ and $$t$$, and the chain rule tells us that

$$
\frac{\partial w}{\partial s} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial s}.
$$

Because $$y = e^t$$ does not depend on $$s$$ (with $$t$$ and $$s$$ independent), we have
$$
\frac{\partial y}{\partial s} = 0.
$$

Thus only the $$x$$-term contributes.

### Step 1. Compute the Partial Derivatives of $$w$$ with Respect to $$x$$ and $$y$$

- Differentiate $$w = y^3 - 5x^2 y$$ with respect to $$x$$:

$$
  \frac{\partial w}{\partial x} = -5\cdot2xy = -10xy.
  $$

- Differentiate $$w$$ with respect to $$y$$:

$$
  \frac{\partial w}{\partial y} = 3y^2 - 5x^2.
  $$

(We will not need $$\frac{\partial w}{\partial y}$$ since $$\frac{\partial y}{\partial s}=0$$.)

### Step 2. Compute $$\frac{\partial x}{\partial s}$$ and $$\frac{\partial y}{\partial s}$$

Since
$$
x = e^{3s},
$$
we have

$$
\frac{\partial x}{\partial s} = 3e^{3s} = 3x.
$$

And, as already noted,

$$
\frac{\partial y}{\partial s} = 0.
$$

### Step 3. Use the Chain Rule

Substitute the derivatives into the chain rule formula:

$$
\frac{\partial w}{\partial s} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial s} = (-10xy)(3x) + \left(3y^2 - 5x^2\right)(0).
$$

This simplifies to

$$
\frac{\partial w}{\partial s} = -30x^2 y.
$$

### Step 4. Evaluate at $$s = -3$$ and $$t = 5$$

First, we find the values of $$x$$ and $$y$$ at these points:

- Since
  $$
  x = e^{3s},
  $$
  at $$s = -3$$:

$$
  x = e^{3(-3)} = e^{-9}.
  $$

- Since
  $$
  y = e^t,
  $$
  at $$t = 5$$:

$$
  y = e^5.
  $$

Now substitute these into the formula for $$\frac{\partial w}{\partial s}$$:

$$
\frac{\partial w}{\partial s} = -30 (e^{-9})^2 (e^5).
$$

Note that $$(e^{-9})^2 = e^{-18}$$, so

$$
\frac{\partial w}{\partial s} = -30 e^{-18} e^5 = -30 e^{-13}.
$$

### Step 5. Compute the Numerical Value

We now approximate $$ e^{-13} $$. Using a calculator or known approximations,

$$
e^{-13} \approx 2.26033\times 10^{-6}.
$$

Thus,

$$
-30 e^{-13} \approx -30 \times 2.26033\times 10^{-6} \approx -6.78\times 10^{-5}.
$$

Rounded to two decimal places (in scientific notation) this is approximately

$$
-6.78 \times 10^{-5}.
$$

Since the value is very small, if one writes it in standard decimal form it would be $$-0.00007$$, which to two decimal places is essentially $$-0.00$$.

### Final Answer

$$
\frac{\partial w}{\partial s} = -30 e^{-13} \approx -6.78\times 10^{-5}.
$$
$$ \frac{\partial w}{\partial s} \approx -0.00 $$

Find using the appropriate Chain Rule for where and , and evaluate the
partial derivative at and . Round your answer to two decimal places.

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Calculus Questions

Find using the appropriate Chain Rule for where and , and evaluate the partial derivative at and . Round your answer to two decimal places.