2 Problems. 1 regarding implicit differentiation

In summary: Then plug it into the equation for the slope of a tangent line. In summary, the first problem asks to find the rate of change of the distance from a moving particle to the origin, as the particle passes through a given point while increasing its x-coordinate at a constant rate. The solution involves using implicit differentiation and the Pythagorean theorem to find the rate of change of the distance. For the second problem, implicit differentiation is used to find the slope of the tangent line to a curve at a given point. The solution involves cross multiplying and plugging in the given coordinates to find the slope.
  • #1
Neil6790
20
0
1)

A particle is moving along the curve y=3sqrt{3x+3}. As the particle passes through the point (2, 9), its x-coordinate increases at a rate of 2 units per second. Find the rate of change of the distance from the particle to the origin at this instant.

2)

Use implicit differentiation to find the slope of the tangent line to the curve
{y}/{x+4y}=x^{3}+9
at the point ( 1,10/-39 ).

The Attempt at a Solution



For the first problem I did it several times and I keep coming up with the wrong answer.

This is my attempt:
dy/dt = 3(1/2)(3x+3)^(-1/2)(3)(dx/dt)
dy/dt = 3/(sqrt(3x+3)) (2)
dy/dt = 3/(sqrt(3(2)+3))
dy/dt = (3/3) * (2/1) = 2
dy/dt = 2

x^2 + y^2 = r^2
2^2 +9^2 = r^2
r = sqrt(85)
2x(dx/dt) + 2y(dy/dt) = 2r(dr/dt)

2(2)(2) + 2(9)(2) = 2(sqrt(85))dr/dt
8+36 = 2(sqrt(85))(dr/dt)
(44/2sqrt(85)) = dr/dt
That is the wrong answer and I have no idea what I am doing wrong.

For the second problem

I did implicit differentiation and came up with

dy/dx = (3x^2(x+4y)^2+y)/x

I plugged in my coordinate values and it was still marked wrong.

Please help me because I really want to understand what I am doing wrong here.

Thanks,
Neil
 
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  • #2
This might help

dr/dt=dr/dx * dx/dt, where dx/dt=2

You are on the right track with x^2 + Y^2= r^2

For part 2 I would do it a little different. I would cross multiply and find dy/dx at point
 

1. What is implicit differentiation?

Implicit differentiation is a mathematical technique used to find the derivative of a function that is not explicitly written in terms of one variable. It is commonly used when the dependent variable cannot be easily isolated on one side of the equation.

2. Why is implicit differentiation useful?

Implicit differentiation allows us to find the rate of change of a function even when it is not written explicitly in terms of one variable. This is useful in cases where the dependent variable is not easily isolated, or when we need to find the derivative of an equation that involves multiple variables.

3. What is the process for implicit differentiation?

The process for implicit differentiation involves differentiating both sides of an equation with respect to the independent variable, using the chain rule and product rule when necessary. The resulting derivative will be in terms of both the dependent and independent variables.

4. Can implicit differentiation be used for any type of equation?

Implicit differentiation can be used for any equation that can be written in the form of F(x,y)=0, where F is a function of x and y. However, it may not always be the most efficient method and other techniques such as the explicit differentiation or logarithmic differentiation may be more appropriate.

5. What are some real-life applications of implicit differentiation?

Implicit differentiation has various applications in physics, economics, and engineering. For example, it can be used to find the marginal cost and revenue in economics, the rate of change of velocity in physics, and the slope of a tangent line in engineering.

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