Determine the value of $b$ in $\frac{dy}{dx} = (bx+3)^3$ given that the tangent to this curve drawn through the point $(1.5, 160)$ also passes through the point $(1, 52)$
So how can I find $b$?
I found out that the equation of the tangent is $y=216x-164$ But how can this help?
Also what confuses me the most is that the question says "tangent to this curve", how can you have a tangent to a curve? I thought you can only have a tangent to a point and not the curve itself?
Assuming the curve is $y=f(x)$ that satisfies the give DE.
Now, given that at the point $(1.5,160)$, we have the tangent equation as
$$\frac{y - 160}{160-52} = \frac{x-1.5}{1.5-1}$$
$$\implies y = 216x -164$$
Hence,
$$\frac{dy}{dx}|_{x=1.5} = 216$$
$$\implies (b(1.5)+3)^3 = 216$$
$$\implies b = 2$$
I assume that you are considering a curve $y(x)$ such that $$\frac{dy}{dx} = (bx+3)^3.$$
Given a curve $y(x)$ and a point $(x_1, y_1)$ (i.e. $y(x_1) = y_1$), the tangent to the curve in that point is a straight line
$$t(x) = mx + q,$$
such that:
$$\begin{cases} t(x_1) = y_1\\ m = \displaystyle\left. \frac{dy}{dx}\right|_{x=x_1} \end{cases}. $$
Now, since you already have find out that $t(x) = 216x - 164,$ the you need to find $b$ such that:
$$m = 216 = \displaystyle\left. \frac{dy}{dx}\right|_{x=x_1} = (b\cdot 1.5 + 3)^3 \Rightarrow b = 2.$$
