Finding the equation of a tangent

Question

I have been asked to find the tangent to the curve $y=\tan(x)$ at the point $(\frac{\pi}{3}, \sqrt{3})$. I am confused because I keep getting a curved line as my answer, but my professors were very clear that the line had to be straight. What is the proper way to solve this?

Answer 1

the line has the equation $$y=mx+n$$ where $m$ is the slope. $$f'(x)=1+\tan^2(x)$$ thus $$f'(\frac{\pi}{3})=1+\tan^2(\frac{\pi}{3})$$ so we have $$y=(1+\tan^2(\frac{\pi}{3})x+n$$ we get the $n$ by plugging the given coordinates of the given Point: $$\sqrt{3}=(1+\tan^2(\frac{\pi}{3}))\frac{\pi}{3}+n$$

Answer 2

You can always write a tangent line in point slope form:

$$(y-y_o) = \frac{dy}{dx}|_{(x_o, y_o)}(x-x_o)$$

where:

• $x,y$ are variables.

• $P(x_o, y_o)$ are coordinates of the point on curve through which tangent passes.

• $\dfrac{dy}{dx} _{(x_o, y_o)}$ is the slope at the point $P(x_o, y_o)$.

One common mistake is that people substitute the whole expression of $y'(x)$ in place of slope. You need the slope at $P(x_o, y_o)$, so you need to substitute $y'(x_o)$ as slope.

So in this manner you always end up with a linear equation in $x,y$ which represents a straight line.

Answer 3

To find the equation of a tangent line you need two things: the slope of the tangent line and a point on the tangent line.

The slope is found by using the derivative. The derivative of $\tan x$ is $\sec^2 x$. Then you need to plug in the given value of $x$ which is $\pi / 3$. So the slope is $(\sec (\pi / 3))^2 = 2^2 = 4$. Many students forget this step and use simply $\sec^2(x)$ as the slope, thus getting a curve instead of a line.

Next, note that we have a point (the point of tangency). It is given to be $(\pi / 3, \sqrt{3})$

Finally, the equation of a line with slope $m$ passing through a point $(x_1, y_1)$ is $y - y_1 = m(x - x_1)$. In this case, we have: $y - \sqrt{3} = 4(x - \pi /3)$.

Answer 4

if $y = tan(x)$ then $\frac{dy}{dx} = sec^2(x)$ and so inputting our value $x = (\frac{π}{3})$ we get $\frac{1}{cos^2(\frac{π}{3})} = 4$

This is the gradient of our tangent so if we substitute into $y-y_1 = m(x-x_1)$ to get $y - \sqrt{3} = 4(x - \frac{π}{3})$

Answer 5

Derivative:

Differentiating $y=\tan x$, we get$$\frac {dy}{dx}=\sec^2x$$ Hence, the slope of the tangent line at any point $x$ is $y_x=\sec^2x$. Now, substituting $x=\tfrac {\pi}3$ into the derivative gives $y_x=4$. So our tangent line takes the form $$4\left(x-\frac {\pi}3\right)=y-\sqrt3$$I’ll let you complete the rest.