Math / Calculus

QuestionSuppose that $w=x^{3} \cdot \exp (2 y) \cdot \cos (7 z)$ with $\begin{array}{c} x=\sin \left(t+\frac{\pi}{2}\right) \\ y=\ln (t+3) \\ z=t \end{array}$ a. Find $\frac{\mathrm{d} w}{\mathrm{~d} t}$ in terms of $t$ . $\frac{\mathrm{d} w}{\mathrm{~d} t}=$

Studdy Solution

STEP 1

What is this asking? We're given a function $w$ that depends on $x$ , $y$ , and $z$ , and each of those variables depends on $t$ .
We need to find how $w$ changes with respect to $t$ , which is $\frac{dw}{dt}$ . Watch out! Don't forget the chain rule!
Since $w$ depends on $x$ , $y$ , and $z$ , and they all depend on $t$ , we need to consider how each variable contributes to the overall change in $w$ .

STEP 2

1. Find the individual derivatives.
2. Apply the chain rule.
3. Substitute and simplify.

STEP 3

Let's start by finding the partial derivatives of $w$ with respect to $x$ , $y$ , and $z$ .
Remember, a partial derivative means we treat the other variables as constants.

STEP 4

First, $\frac{\partial w}{\partial x}$ .
Treating $y$ and $z$ as constants, we get: $\frac{\partial w}{\partial x} = 3x^2 \cdot \exp(2y) \cdot \cos(7z)$ We brought down the exponent of $x$ and reduced it by one, just like regular differentiation.

STEP 5

Next, $\frac{\partial w}{\partial y}$ : $\frac{\partial w}{\partial y} = x^3 \cdot 2\exp(2y) \cdot \cos(7z)$ The derivative of $\exp(2y)$ with respect to $y$ is $2\exp(2y)$ .

STEP 6

Finally, $\frac{\partial w}{\partial z}$ : $\frac{\partial w}{\partial z} = x^3 \cdot \exp(2y) \cdot (-7\sin(7z))$ The derivative of $\cos(7z)$ with respect to $z$ is $-7\sin(7z)$ .

STEP 7

Now, let's find the derivatives of $x$ , $y$ , and $z$ with respect to $t$ : $\frac{dx}{dt} = \cos\left(t + \frac{\pi}{2}\right)$ $\frac{dy}{dt} = \frac{1}{t+3}$ $\frac{dz}{dt} = 1$

STEP 8

The chain rule tells us that: $\frac{dw}{dt} = \frac{\partial w}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial w}{\partial y} \cdot \frac{dy}{dt} + \frac{\partial w}{\partial z} \cdot \frac{dz}{dt}$ This formula shows how the change in each variable $x$ , $y$ , and $z$ contributes to the overall change in $w$ .

STEP 9

Now we substitute all our derivatives into the chain rule formula: $\frac{dw}{dt} = \left( 3x^2 \cdot \exp(2y) \cdot \cos(7z) \right) \cdot \cos\left(t + \frac{\pi}{2}\right) + \left( x^3 \cdot 2\exp(2y) \cdot \cos(7z) \right) \cdot \frac{1}{t+3} + \left( x^3 \cdot \exp(2y) \cdot (-7\sin(7z)) \right) \cdot 1$

STEP 10

Let's substitute the expressions for $x$ , $y$ , and $z$ in terms of $t$ : $\frac{dw}{dt} = 3\sin^2\left(t+\frac{\pi}{2}\right)\exp(2\ln(t+3))\cos(7t)\cos\left(t+\frac{\pi}{2}\right) + \frac{2\sin^3\left(t+\frac{\pi}{2}\right)\exp(2\ln(t+3))\cos(7t)}{t+3} - 7\sin^3\left(t+\frac{\pi}{2}\right)\exp(2\ln(t+3))\sin(7t)$

STEP 11

We can simplify this using the fact that $\exp(2\ln(t+3)) = (t+3)^2$ : $\frac{dw}{dt} = 3(t+3)^2\sin^2\left(t+\frac{\pi}{2}\right)\cos(7t)\cos\left(t+\frac{\pi}{2}\right) + 2(t+3)\sin^3\left(t+\frac{\pi}{2}\right)\cos(7t) - 7(t+3)^2\sin^3\left(t+\frac{\pi}{2}\right)\sin(7t)$

STEP 12

$\frac{dw}{dt} = 3(t+3)^2\sin^2\left(t+\frac{\pi}{2}\right)\cos(7t)\cos\left(t+\frac{\pi}{2}\right) + 2(t+3)\sin^3\left(t+\frac{\pi}{2}\right)\cos(7t) - 7(t+3)^2\sin^3\left(t+\frac{\pi}{2}\right)\sin(7t)$

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QuestionSuppose that $w=x^{3} \cdot \exp (2 y) \cdot \cos (7 z)$ with $\begin{array}{c} x=\sin \left(t+\frac{\pi}{2}\right) \\ y=\ln (t+3) \\ z=t \end{array}$ a. Find $\frac{\mathrm{d} w}{\mathrm{~d} t}$ in terms of $t$ . $\frac{\mathrm{d} w}{\mathrm{~d} t}=$

Studdy Solution

STEP 1

STEP 2

1. Find the individual derivatives.
2. Apply the chain rule.
3. Substitute and simplify.

STEP 3

Let's start by finding the partial derivatives of $w$ with respect to $x$ , $y$ , and $z$ .
Remember, a partial derivative means we treat the other variables as constants.

STEP 4

First, $\frac{\partial w}{\partial x}$ .
Treating $y$ and $z$ as constants, we get: $\frac{\partial w}{\partial x} = 3x^2 \cdot \exp(2y) \cdot \cos(7z)$ We brought down the exponent of $x$ and reduced it by one, just like regular differentiation.

STEP 5

Next, $\frac{\partial w}{\partial y}$ : $\frac{\partial w}{\partial y} = x^3 \cdot 2\exp(2y) \cdot \cos(7z)$ The derivative of $\exp(2y)$ with respect to $y$ is $2\exp(2y)$ .

STEP 6

Finally, $\frac{\partial w}{\partial z}$ : $\frac{\partial w}{\partial z} = x^3 \cdot \exp(2y) \cdot (-7\sin(7z))$ The derivative of $\cos(7z)$ with respect to $z$ is $-7\sin(7z)$ .

STEP 7

Now, let's find the derivatives of $x$ , $y$ , and $z$ with respect to $t$ : $\frac{dx}{dt} = \cos\left(t + \frac{\pi}{2}\right)$ $\frac{dy}{dt} = \frac{1}{t+3}$ $\frac{dz}{dt} = 1$

STEP 8

STEP 9

STEP 10

STEP 11

We can simplify this using the fact that $\exp(2\ln(t+3)) = (t+3)^2$ : $\frac{dw}{dt} = 3(t+3)^2\sin^2\left(t+\frac{\pi}{2}\right)\cos(7t)\cos\left(t+\frac{\pi}{2}\right) + 2(t+3)\sin^3\left(t+\frac{\pi}{2}\right)\cos(7t) - 7(t+3)^2\sin^3\left(t+\frac{\pi}{2}\right)\sin(7t)$

STEP 12

$\frac{dw}{dt} = 3(t+3)^2\sin^2\left(t+\frac{\pi}{2}\right)\cos(7t)\cos\left(t+\frac{\pi}{2}\right) + 2(t+3)\sin^3\left(t+\frac{\pi}{2}\right)\cos(7t) - 7(t+3)^2\sin^3\left(t+\frac{\pi}{2}\right)\sin(7t)$

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Studdy Solution

STEP 1

STEP 2

1. Find the individual derivatives.2. Apply the chain rule.3. Substitute and simplify.

STEP 3

Let's **start** by finding the partial derivatives of www with respect to xxx, yyy, and zzz.Remember, a partial derivative means we treat the *other* variables as constants.

STEP 4

STEP 5

Next, ∂w∂y\frac{\partial w}{\partial y}∂y∂w​: ∂w∂y=x3⋅2exp⁡(2y)⋅cos⁡(7z) \frac{\partial w}{\partial y} = x^3 \cdot 2\exp(2y) \cdot \cos(7z) ∂y∂w​=x3⋅2exp(2y)⋅cos(7z) The derivative of exp⁡(2y)\exp(2y)exp(2y) with respect to yyy is 2exp⁡(2y)2\exp(2y)2exp(2y).

STEP 6

STEP 7

Now, let's find the derivatives of xxx, yyy, and zzz with respect to ttt: dxdt=cos⁡(t+π2) \frac{dx}{dt} = \cos\left(t + \frac{\pi}{2}\right) dtdx​=cos(t+2π​) dydt=1t+3 \frac{dy}{dt} = \frac{1}{t+3} dtdy​=t+31​dzdt=1 \frac{dz}{dt} = 1 dtdz​=1

STEP 8

STEP 9

STEP 10

STEP 11

STEP 12

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1. Find the individual derivatives.
2. Apply the chain rule.
3. Substitute and simplify.

Let's start by finding the partial derivatives of $w$ with respect to $x$ , $y$ , and $z$ .
Remember, a partial derivative means we treat the other variables as constants.

Next, $\frac{\partial w}{\partial y}$ : $\frac{\partial w}{\partial y} = x^3 \cdot 2\exp(2y) \cdot \cos(7z)$ The derivative of $\exp(2y)$ with respect to $y$ is $2\exp(2y)$ .

Now, let's find the derivatives of $x$ , $y$ , and $z$ with respect to $t$ : $\frac{dx}{dt} = \cos\left(t + \frac{\pi}{2}\right)$ $\frac{dy}{dt} = \frac{1}{t+3}$ $\frac{dz}{dt} = 1$