Solve the following equation for y: 35x-7y=49

Use chain rule to find dw/dt w=ln(x^2 + y^2 + z^2)^(1/2) , x=sint, y=cost, z=tant

son4ous [18]

The chain rule states that

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\partial w}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial w}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}+\dfrac{\partial w}{\partial z}\dfrac{\mathrm dz}{\mathrm dt}$

Since $\ln(x^2+y^2+z^2)^{1/2}=\dfrac12\ln(x^2+y^2+z^2)$ , you have

$\dfrac{\partial w}{\partial x}=\dfrac x{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial x}=\dfrac y{x^2+y^2+z^2}$
$\dfrac{\partial w}{\partial z}=\dfrac z{x^2+y^2+z^2}$

and

$\dfrac{\mathrm dx}{\mathrm dt}=\cos t$
$\dfrac{\mathrm dy}{\mathrm dt}=-\sin t$
$\dfrac{\mathrm dz}{\mathrm dt}=\sec^2t$

Also, since $x^2+y^2+z^2=\sin^2t+\cos^2t+\tan^2t=1+\tan^2t=\sec^2t$ , the derivative is

$\dfrac{\mathrm dw}{\mathrm dt}=\dfrac{\sin t\cos t}{\sec^2t}-\dfrac{\sin t\cos t}{\sec^2t}+\dfrac{\tan t\sec^2t}{\sec^2t}$
$\dfrac{\mathrm dw}{\mathrm dt}=\tan t$

4 0

2 years ago

Substitute each variable to determine whether the inequality is true or false.

algol13

Answer:

It is false: -16>-15 is not true

3 0

2 years ago

Read 2 more answers

Plzz help this is due today

andreyandreev [35.5K]

Answer:

Major Arc

Step-by-step explanation:

because it's bigger then half of the circle

8 0

2 years ago

Read 2 more answers

If <3=70°, then find the rest of the angles.

Sati [7]

If <3=70°, then find the rest of the angles.

3/4

5/6

7/8

79

1.) m<1=

2.) m<2=

4.) m<5=

3.) m<4=

6.) m<7=

5.) m<6=

subscribes to iwpp

6 0

2 years ago

A car passes a landmark traveling at a constant rate of 55 kilometers per hour. One hour later, a second car passes the same lan

yarga [219]

Recall your d = rt, distance = rate * time

one car is going at 55kph, at say hmmm "t" time
the second car zooms by at 75kph, an hour later, thus " t + 1 "

notice, overtaking the 1st car, simply means, the second car comes from behind approaches the first car from behind and "gets there", at that very second, both cars have travelled a distance, say "d"

if the first car travelled "d" kilometers
for the second car to "get there" has to had travelled "d" distance as well

$\bf \begin{array}{lccclll}
&distance&rate&time\\
&-----&-----&-----\\
\textit{first car}&d&55&t\\
\textit{second car}&d&75&t+1
\end{array}
\\\\\\
\begin{cases}
d=55t\\
d=75(t+1)\\
-------\\
55t=75(t+1)
\end{cases}$

solve for "t"

7 0

2 years ago

Read 2 more answers

Solve the following equation for y: 35x-7y=49​

Solve the following equation for y: 35x-7y=49