find the equation of the tangent line to y^3=x^2 if that is parallel to the line y=4x-1

Question

alenahelenash · Accepted Answer

To find the equation of the tangent line to the curve

y^{3} = x^{2}

that is parallel to the line

y = 4 x - 1

, we can follow these steps:
First, let's differentiate the equation

y^{3} = x^{2}

implicitly with respect to

x

to find the derivative of

y

with respect to

x

:

\frac{d}{d x} (y^{3}) = \frac{d}{d x} (x^{2})

Using the chain rule, we have:

3 y^{2} \cdot \frac{d y}{d x} = 2 x

Next, let's find the slope of the tangent line by setting the derivative equal to the slope of the given line

y = 4 x - 1

:

3 y^{2} \cdot \frac{d y}{d x} = 2 x = 4

Simplifying, we get:

3 y^{2} \cdot \frac{d y}{d x} = 4

Now, we need to find the values of

x

and

y

that satisfy both equations simultaneously. Since the tangent line is parallel to

y = 4 x - 1

, its slope must be

4

. Therefore, we have:

3 y^{2} \cdot \frac{d y}{d x} = 4

Since we want the tangent line to be parallel, we can substitute

4

for

\frac{d y}{d x}

:

3 y^{2} \cdot 4 = 4

Simplifying further, we get:

12 y^{2} = 4

Dividing both sides by

12

, we have:

y^{2} = \frac{1}{3}

Taking the square root of both sides, we find two possible values for

y

:

y = \pm \frac{1}{\sqrt{3}}

Now, let's find the corresponding values of

x

for these

y

values. Substituting

y = \frac{1}{\sqrt{3}}

into the equation

y^{3} = x^{2}

, we have:

{(\frac{1}{\sqrt{3}})}^{3} = x^{2}

Simplifying, we get:

\frac{1}{3 \sqrt{3}} = x^{2}

Taking the square root of both sides, we find:

x = \pm \frac{1}{\sqrt{3} \sqrt{\sqrt{3}}} = \pm \frac{1}{\sqrt{3} \cdot \sqrt[4]{3}}

Therefore, we have two points that satisfy both equations:

(\frac{1}{\sqrt{3} \cdot \sqrt[4]{3}}, \frac{1}{\sqrt{3}})

and

(- \frac{1}{\sqrt{3} \cdot \sqrt[4]{3}}, \frac{1}{\sqrt{3}})

.
Now, we can use the point-slope form of a linear equation to find the equation of the tangent line. We'll use the point

(\frac{1}{\sqrt{3} \cdot \sqrt[4]{3}}, \frac{1}{\sqrt{3}})

.
The slope of the tangent line is

4

, so the equation of the tangent line is:

y - \frac{1}{\sqrt{3}} = 4 (x - \frac{1}{\sqrt{3} \cdot \sqrt[4]{3}})

Simplifying and rearranging, we get:

y = 4 x - \frac{4}{\sqrt{3}} + \frac{1}{\sqrt{3}}

Combining the constants, we have:

y = 4 x - \frac{3}{\sqrt{3}}

Finally, we can rationalize the denominator by multiplying the numerator and denominator by

\sqrt{3}

:

y = 4 x - \sqrt{3}

Therefore, the equation of the tangent line to

y^{3} = x^{2}

that is parallel to

y = 4 x - 1

is

y = 4 x - \sqrt{3}

.

find the equation of the tangent line to

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