# Tangent line Question

I am stumped with this question.

Find the line tangent $$f(t)=3\sin(2t)+5$$ at the point where $$t=\pi$$.

You must first find the derivative of $$f$$ at $$\pi$$. Next, find the equation of the tangent line using the slope$$=f′(π)$$ and the point $$P(\pi)$$. Yep, you need to find $$y$$.

• Hello and welcome to Mathematics StackExchange! Do you know the equation for the tangent line of a function at some point? There is a general formula for that. – Markus Zetto May 5 '20 at 16:08

The problem specifies the point $$(t_1,y_1)=\left(\pi,f(\pi)\right)$$ where $$f(t)=3\sin(2t)+5$$ and $$y=f(t)$$. The equation of a tangent line is given by

$$y-y_1=m(t-t_1).\tag{1}$$

You need to find $$f(\pi)$$ and $$m=f'(\pi)$$ and then substitute them into $$(1)$$.

Step 1: Find the derivative of $$f$$ at $$\pi$$.

$$f(t)=3\sin(2t)+5$$ $$f'(t)=6\cos(2t)$$ $$f'(\pi)=6\cos(2\pi)=6$$

Step 2: Find $$f(\pi)$$.

$$f(\pi)=3\sin(2\pi)+5=5$$

Therefore, the point is $$(t_1,y_1)=(\pi,5)$$ and the slope of the tangent line is $$m=f'(\pi)=6$$. Substitute these values into $$(1)$$ to find the equation of the tangent line.

The derivative is the slope of the tangent line. If you plug in pi into the equation of the derivative you get the slope at the point pi: f’( pi ). If you have a point and a slope, you can use the point-slope formula to express a line. If you plug in pi in to the equation, you get the point ( pi, f( pi ) ).

At $$f'(π)=6cos(2π)=6$$. So, the slope is positive.

So we get that the equation of the tangent line will be

$$\frac{y-f(π)}{x-π} =6 \Rightarrow y=6x+(5-6π)$$.