Question

Find an equation of the tangent line to the curve y=8x sinx at the point pi/2 4pi

Answer 1

Hello
We know that upon differentiating a function at a certain point, we can obtain its gradient at that point. That will also be the gradient of its tangent line. Thus,
y = 8x * sin(x)
y' = 8x * cos(x) + 8sin(x)
Evaluating the gradient at the given point, we insert pi/2 for x.
gradient = 8(pi/2) * cos(pi/2) + 8sin(pi/2)
= 2pi*sqrt(2) + 4(sqrt(2))
The equation of the line will be in the form y = mx + c; where m is the gradient and c is the y intercept. To calculate the y-int, we insert the given x and y values:
4pi = [2pi*sqrt(2)+4(sqrt(2))]*(pi/2) + c
c = -1.98
Thus, the equation is:
y = 14.5x - 1.98

Answer 2

The equation of tangent line is $\boxed{y=8x}$.

Further explanation:

A tangent line is a line which touches a curve at one and only one point.

The equation of a tangent line is $y=mx+c$ where $m$ is the slope of the line is and $c$ is the $y$-intercept.

Calculation:

The given curve is $y=8x\text{sin}x$.

Differentiate the above curve using product rule of differentiation as follows:

$\begin{aligned}y'&=\dfrac{d}{dx}(8x\text{sin}x)\\&=8\text{sin}x+8x\text{cos}x\end{aligned}$

Label the above equation as follows:

$\boxed{y'=8\text{sin}x+8x\text{cos}x}$   .......(1)

The point on the curve through which the tangent line passes is $\left(\dfrac{\pi}{2},4\pi\right)$.

Substitute $x=\frac{\pi}{2}$ in equation (1).

$\begin{aligned}y'&=8\text{sin}\left(\dfrac{\pi}{2}\right)+\left(8\cdot \dfrac{\pi}{2}\cdot \text{cos}\left(\dfrac{\pi}{2}\right)\right)\\&=8\cdot 1+\left(8\cdot \dfrac{\pi}{2}\cdot 0\right)\\&=8\end{aligned}$

Therefore, the slope of the tangent line is $8$.

The equation of the tangent line is $y=mx+c$ where $m$ is the slope and $c$ is the $y$-intercept.

Now, substitute the value of $y,m\text{ and }x$ in the equation of tangent line to obtain the value of $c$ as follows:

$\begin{aligned}4\pi&=8\cdot \left(\dfrac{\pi}{2}\right)\\ \4\pi&=4\pi +c\\c&=0\end{aligned}$

The $\bf y$ intercept of the tangent line is $c=0$.

Now, substitute the values of $m$ and $c$ in $y=mx+c$.

$\begin{aligned}y&=8x+0\\y&=8x\end{aligned}$

Therefore, the equation of tangent is $\boxed{y=8x}$.  

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Answer details:

Grade: High school

Subject: Mathematics

Chapter: Lines and Tangents

Keywords: Tangents, equation of tangent line, slope, intercept, curve, line, differentiation, line, calculus, y=8xsinx, pi/2, y-intercept, derivative, product rule, derivative.