Here's your answers! Hope this helps.
Answer:
The rate of change of the distance
when x = 9 and y = 12 is
.
Step-by-step explanation:
This is an example of a related rate problem. A related rate problem is a problem in which we know one of the rates of change at a given instant
and we want to find the other rate
at that instant.
We know the rate of change of x-coordinate and y-coordinate:
![\frac{dx}{dt}=-2\:\frac{m}{s} \\\\\frac{dy}{dt}=-8\:\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bdx%7D%7Bdt%7D%3D-2%5C%3A%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5C%5C%5C%5Cfrac%7Bdy%7D%7Bdt%7D%3D-8%5C%3A%5Cfrac%7Bm%7D%7Bs%7D)
We want to find the rate of change of the distance
when x = 9 and y = 12.
The distance of a point (x, y) and the origin is calculated by:
![s=\sqrt{x^2+y^2}](https://tex.z-dn.net/?f=s%3D%5Csqrt%7Bx%5E2%2By%5E2%7D)
We need to use the concept of implicit differentiation, we differentiate each side of an equation with two variables by treating one of the variables as a function of the other.
If we apply implicit differentiation in the formula of the distance we get
![s=\sqrt{x^2+y^2}\\\\s^2=x^2+y^2\\\\2s\frac{ds}{dt}= 2x\frac{dx}{dt}+2y\frac{dy}{dt}\\\\\frac{ds}{dt}=\frac{1}{s}(x\frac{dx}{dt}+y\frac{dy}{dt})](https://tex.z-dn.net/?f=s%3D%5Csqrt%7Bx%5E2%2By%5E2%7D%5C%5C%5C%5Cs%5E2%3Dx%5E2%2By%5E2%5C%5C%5C%5C2s%5Cfrac%7Bds%7D%7Bdt%7D%3D%202x%5Cfrac%7Bdx%7D%7Bdt%7D%2B2y%5Cfrac%7Bdy%7D%7Bdt%7D%5C%5C%5C%5C%5Cfrac%7Bds%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7Bs%7D%28x%5Cfrac%7Bdx%7D%7Bdt%7D%2By%5Cfrac%7Bdy%7D%7Bdt%7D%29)
Substituting the values we know into the above formula
![s=\sqrt{9^2+12^2}=15](https://tex.z-dn.net/?f=s%3D%5Csqrt%7B9%5E2%2B12%5E2%7D%3D15)
![\frac{ds}{dt}=\frac{1}{15}(9(-2)+12(-8))\\\\\frac{ds}{dt}=\frac{1}{15}\left(-18-96\right)\\\\\frac{ds}{dt}=\frac{1}{15}(-114)=-\frac{38}{5}=-7.6\:\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bds%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B15%7D%289%28-2%29%2B12%28-8%29%29%5C%5C%5C%5C%5Cfrac%7Bds%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B15%7D%5Cleft%28-18-96%5Cright%29%5C%5C%5C%5C%5Cfrac%7Bds%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B15%7D%28-114%29%3D-%5Cfrac%7B38%7D%7B5%7D%3D-7.6%5C%3A%5Cfrac%7Bm%7D%7Bs%7D)
The rate of change of the distance
when x = 9 and y = 12 is ![-7.6\:\frac{m}{s}](https://tex.z-dn.net/?f=-7.6%5C%3A%5Cfrac%7Bm%7D%7Bs%7D)
Answer:
![\fbox{Gradient of curve = -6}](https://tex.z-dn.net/?f=%5Cfbox%7BGradient%20of%20curve%20%3D%20-6%7D)
<em><u>Step by step explanation:</u></em>
<em>Given</em>:
Equation of curve ↦ xy = 6
Point ↦(1,6)
<em>To find:</em>
<em>The gradient of curve = ?</em>
Solution:
![Using \: the \: equation, \\ xy = 6 \\](https://tex.z-dn.net/?f=Using%20%5C%3A%20%20the%20%5C%3A%20%20equation%2C%20%5C%5C%20xy%20%3D%206%20%5C%5C%20)
Taking log both the side,
![log(xy) = log(6) \\ log(x) + log(y) = log(6) \\ Taking \: derivative \: both \: the \: side \\ \frac{1}{x} + \frac{1}{y} \cdot\frac{dy}{dx} = \: 0 \\ (Derivative \: of \: any \: constant =0) \\ \frac{1}{y} \cdot\frac{dy}{dx} = - \frac{1}{x} \\ \frac{dy}{dx} = - \frac{y}{x} \\ To \: find \: gradient \: at \: point (1,6) \\ \frac{dy}{dx} \mid_{(1,6)} = \frac{ - 6}{1} \\ \frac{dy}{dx} \mid_{(1,6)} = - 6 \\ \fbox{Gradient of curve = -6}](https://tex.z-dn.net/?f=log%28xy%29%20%3D%20log%286%29%20%5C%5C%20log%28x%29%20%2B%20log%28y%29%20%3D%20log%286%29%20%5C%5C%20Taking%20%5C%3A%20%20derivative%20%5C%3A%20%20both%20%20%5C%3A%20the%20%20%5C%3A%20side%20%20%5C%5C%20%20%5Cfrac%7B1%7D%7Bx%7D%20%20%2B%20%20%5Cfrac%7B1%7D%7By%7D%20%20%5Ccdot%5Cfrac%7Bdy%7D%7Bdx%7D%20%20%3D%20%20%5C%3A%200%20%5C%5C%20%28Derivative%20%20%5C%3A%20of%20%5C%3A%20%20any%20%5C%3A%20%20constant%20%3D0%29%20%5C%5C%20%5Cfrac%7B1%7D%7By%7D%20%20%5Ccdot%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%20-%20%5Cfrac%7B1%7D%7Bx%7D%20%5C%5C%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%20%3D%20-%20%5Cfrac%7By%7D%7Bx%7D%20%5C%5C%20To%20%20%5C%3A%20find%20%20%5C%3A%20gradient%20%20%5C%3A%20at%20%20%5C%3A%20point%20%281%2C6%29%20%5C%5C%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cmid_%7B%281%2C6%29%7D%20%3D%20%20%5Cfrac%7B%20-%206%7D%7B1%7D%20%20%5C%5C%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cmid_%7B%281%2C6%29%7D%20%3D%20%20-%206%20%5C%5C%20%5Cfbox%7BGradient%20of%20curve%20%3D%20-6%7D)
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Answer:
2,144.66 in³
Step-by-step explanation:
Thus, the volume of the water in the fishbowl = volume of a sphere having diameter of 16 inches
= 2,144.66 in³
Answer:it would be 12
Step-by-step explanation:
Well it would be 12 because ac is half of 24 which would be dividing by 2 which is why u would have 12 as the radius.