# Calculus III: Proving Tangent Line = 1 for Every Point on Curve

• MHB
• Cactusguy21
In summary, you need to find the equation of the tangent line, solve for $y-y_0$, and find the point where it crosses the $x$-axis and the $y$-axis.
Cactusguy21
View attachment 8471

Hi, I'm stuck on a homework problem in my Calculus III class.

I solved 3a really easily, but 3b is giving me a lot of trouble. I know that to find the tangent line, I first have to find the slope, which is represented by the vector:

<3cos^2(t)(-sin(t)), 3sin^2(t)(cos(t))>.
I know the formula for arc length as well, although I don't think I necessarily would need this.

The formula for the tangent line should be:

r'(t)= <cos^3(t), sin^3(t)> + s<3cos^2(t)(-sin(t)), 3sin^2(t)(cos(t))>
where s is a parameter.

But how do I set it up so that I can prove the rest of the problem?

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Cactusguy21 said:
Hi, I'm stuck on a homework problem in my Calculus III class.

I solved 3a really easily, but 3b is giving me a lot of trouble. I know that to find the tangent line, I first have to find the slope, which is represented by the vector:

<3cos^2(t)(-sin(t)), 3sin^2(t)(cos(t))>.
I know the formula for arc length as well, although I don't think I necessarily would need this.

The formula for the tangent line should be:

r'(t)= <cos^3(t), sin^3(t)> + s<3cos^2(t)(-sin(t)), 3sin^2(t)(cos(t))>
where s is a parameter.

But how do I set it up so that I can prove the rest of the problem?
Hi Cactusguy, and welcome to MHB!

What you have done so far is correct. Now find the equation of the tangent line in the form $y-y_0 = m(x-x_0)$, where $(x_0,y_0)$ is the point $(cos^3t,sin^3t)$ on the line, and $m$ is the slope, which from your calculation is $\dfrac{3\sin^2t\cos t}{-3\cos^2t(-\sin t)} = -\dfrac{\sin t}{\cos t}$. So the equation of the tangent line is $y-sin^3t = -\frac{\sin t}{\cos t}(x - \cos^3t).$

Simplify that equation, then put $y=0$ to find where it crosses the $x$-axis, put $x=0$ to find where it crosses the $y$-axis, and finally compute the distance between those points.

## 1. What is Calculus III and why is it important?

Calculus III is the third course in the calculus sequence and it builds upon the concepts learned in Calculus I and II. It focuses on multivariable calculus, which is used to study functions of multiple variables. This course is important because it is a fundamental tool in many fields of science and engineering, and it allows for a deeper understanding of real-world problems.

## 2. What does it mean to prove that the tangent line equals 1 for every point on a curve?

Proving that the tangent line equals 1 for every point on a curve means that at any given point on the curve, the slope of the tangent line is equal to 1. This can be thought of as the curve being "flat" at every point, with a slope of 1.

## 3. How is this proof different from other proofs in Calculus III?

This proof is different from other proofs in Calculus III because it specifically focuses on proving a property of tangent lines on a curve. Other proofs in Calculus III may involve different concepts and properties, such as integration, differentiation, and optimization.

## 4. What are some real-world applications of proving the tangent line equals 1 for every point on a curve?

Some real-world applications of this proof include optimization problems in physics and engineering, such as finding the minimum or maximum value of a function, and in economics, where slope is often used to represent the rate of change in a particular situation.

## 5. How is this proof used in other fields of science?

This proof is used in other fields of science, such as physics, engineering, and economics, as it provides a fundamental understanding of tangent lines and their properties. It can also be applied to other areas of mathematics, such as geometry and differential equations.

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