Question 5
The gradient of the tangent at point A on the curve $y^2=4x$ is $1.36$.
The $x$-coordinate of A is
Not quite. $0.12$ comes from setting $2y=1.36$, as if the gradient were the derivative of $y^2$ on its own. That gives $y=0.68$ and $x=\frac{0.68^2}{4}\approx0.12$. Differentiating both sides gives $2y\frac{dy}{dx}=4$, so $\frac{dy}{dx}=\frac{2}{y}$.
Not quite. $0.46$ uses $1.36$ as the $y$-coordinate: $x=\frac{1.36^2}{4}\approx0.46$. The gradient there is $\frac{2}{1.36}\approx1.47$. Set the gradient $\frac{2}{y}=1.36$ to find $y$ first.
That’s it. Step through the hints below to see why.
Not quite. $1.47$ is the $y$-coordinate of A: $\frac{2}{y}=1.36$ gives $y\approx1.4706$. The question asks for $x$, so finish with $x=\frac{y^2}{4}\approx0.54$.
Both axes use the same scale, so the steepness of the orange tangent is its true gradient. A positive gradient puts A on the upper half of the curve.
The tangent at A has gradient about $1.36$. This is option (C). The unrounded value is $x\approx0.5407$.
At $x=0.12$ the gradient is about $2.89$, far steeper than $1.36$. Here $y\approx0.69$, close to the $0.68$ you get from $2y=1.36$.
At $x=0.46$ the $y$-coordinate is about $1.36$. That is the height of the point. The gradient here is about $1.47$.
At $x=1.47$ the gradient is only about $0.82$. The number $1.47$ is the $y$-coordinate of the point you want, which sits at $x\approx0.54$.
The tangent is steeper than $1.36$ here. Move A to the right.
The tangent is less steep than $1.36$ here. Move A to the left.
Differentiate both sides with respect to $x$. The chain rule turns $y^2$ into $2y\frac{dy}{dx}$.
$$2y\frac{dy}{dx}=4\ \Rightarrow\ \frac{dy}{dx}=\frac{2}{y}$$
Set the gradient equal to $1.36$ and solve for $y$. The gradient is positive, so A is on the upper half.
$$\frac{2}{y}=1.36\ \Rightarrow\ y=\frac{2}{1.36}\approx1.4706$$
Substitute into the curve to find $x$.
$$x=\frac{y^2}{4}\approx\frac{1.4706^2}{4}\approx0.5407$$
Differentiate implicitly.
$$2y\frac{dy}{dx}=4\ \Rightarrow\ \frac{dy}{dx}=\frac{2}{y}$$
$$\frac{2}{y}=1.36\ \Rightarrow\ y\approx1.4706$$
$$x=\frac{y^2}{4}\approx0.5407$$
The $x$-coordinate of A is about $0.54$.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
Spot a mistake on this page? Report it