# Using dx/dy to find a y-parallel tangent

• Faiq
In summary: Faiq, in the future, please post questions about derivatives in the Calculus & Beyond section, not the Precalc section.
Faiq

## Homework Statement

The curve ##C## has equation
$$y=x^2+0.2sin(x+y)$$
Show that ##C## has no tangent(no point where ##dy/dx=∞##), that is parallel to the y axis.

Attempt
$$1=2x\frac{dx}{dy}+0.2cos(x+y)(1+\frac{dx}{dy})$$

For a tangent to be parallel to y-axis,

$$\frac{dx}{dy}=0$$

$$1=0.2cos(x+y)$$

$$cos(x+y)=5$$

No value of ##(x,y)## exist for which ##cos(x+y)>1##, hence no y-axis parallel tangent.

**Confusion**
Never been taught or encountered ##\frac{dx}{dy}## and thus I am having doubts over the validity of the solution.

What about ##dx\over dy## if ##{dy\over dx} < Z## where you can show that Z is finite ?

Thread moved. @Faiq, in the future, please post questions about derivatives in the Calculus & Beyond section, not the Precalc section.

Why not just find dy/dx, and show that it is defined for all real x?

BvU said:
What about ##dx\over dy## if ##{dy\over dx} < Z## where you can show that Z is finite ?
OK, ' is finite for all finite x ' -- thanks, Mark !

BvU said:
What about ##dx\over dy## if ##{dy\over dx} < Z## where you can show that Z is finite ?
Not sure what you're telling. Can you please elaborate?

Mark44 said:
Why not just find dy/dx, and show that it is defined for all real x?
I am just concerned whether my solution is correct or not.

Faiq said:
I am just concerned whether my solution is correct or not.
Your solution looks fine to me.

Faiq said:

## Homework Statement

The curve ##C## has equation
$$y=x^2+0.2sin(x+y)$$
Show that ##C## has no tangent(no point where ##dy/dx=∞##), that is parallel to the y axis.

Attempt
$$1=2x\frac{dx}{dy}+0.2cos(x+y)(1+\frac{dx}{dy})$$

For a tangent to be parallel to y-axis,

$$\frac{dx}{dy}=0$$

$$1=0.2cos(x+y)$$

$$cos(x+y)=5$$

No value of ##(x,y)## exist for which ##cos(x+y)>1##, hence no y-axis parallel tangent.

**Confusion**
Never been taught or encountered ##\frac{dx}{dy}## and thus I am having doubts over the validity of the solution.

I guess it is correct because by inverse function theorem ##D_x (y) = 1/D_y(x)##. So if ##D_y(x) \to \infty##, ## D_x(y) \to 0##.

Faiq said:
I am just concerned whether my solution is correct or not.

In that case, you may need to show how it is that requiring dx/dy = 0 gets you from
##\displaystyle 1=2x\frac{dx}{dy}+0.2cos(x+y)(1+\frac{dx}{dy}) ##​
to
##\displaystyle 1=0.2cos(x+y) ##​
.

Okay thanks to all of you very much.

## 1. How is dx/dy used to find a y-parallel tangent?

To find a y-parallel tangent using dx/dy, you first need to find the derivative of the function at the given point. This derivative, dx/dy, represents the slope of the tangent line at that point. Then, you can use this slope and the given point to create an equation for the tangent line, which will be parallel to the y-axis.

## 2. What is the significance of using dx/dy to find a y-parallel tangent?

Using dx/dy to find a y-parallel tangent allows you to determine the slope of the tangent line at a specific point on a curve. This can be helpful in understanding the behavior of the curve and making predictions about its behavior in the future.

## 3. Can dx/dy be used to find a y-parallel tangent for any function?

Yes, dx/dy can be used to find a y-parallel tangent for any differentiable function. This means that the function must have a defined derivative at the given point in order for dx/dy to be used.

## 4. How do you know if a y-parallel tangent exists for a given point on a curve?

A y-parallel tangent will exist for any point on a curve where the derivative of the function at that point is defined. This means that the function must have a non-vertical tangent at that point in order for a y-parallel tangent to exist.

## 5. Can you use dx/dy to find a y-parallel tangent at a point where the function is not differentiable?

No, dx/dy can only be used to find a y-parallel tangent at points where the function is differentiable. If the function is not differentiable at a given point, then the slope of the tangent line at that point is undefined and dx/dy cannot be used.

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