A curve in the plane is defined parametrically by the equations $\newline$ $x=t^{3}+t$ and $\newline$ $y=t^{4}+2t^{2}$ . An equation of the line tangent to the curve at $\newline$ $t=1$ is $\newline$ $\text{(A)}\ y=2x$ $\newline$ $\text{(B)}\ y=8x$ $\newline$ $\text{(C)}\ y=2x-1$ $\newline$ $\text{(D)}\ y=4x-5$ $\newline$ $\text{(E)}\ y=8x+13$

View step-by-step help

Home

Math Problems

Precalculus

Find equations of tangent lines using limits

Full solution

Q. A curve in the plane is defined parametrically by the equations

\newline

x=t^{3}+t

and

\newline

y=t^{4}+2t^{2}

. An equation of the line tangent to the curve at

\newline

t=1

\newline

\text{(A)}\ y=2x

\newline

\text{(B)}\ y=8x

\newline

\text{(C)}\ y=2x-1

\newline

\text{(D)}\ y=4x-5

\newline

\text{(E)}\ y=8x+13

Find Derivatives: To find the equation of the tangent line to the curve at a specific point, we first need to find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$ , which will give us the slope of the tangent line at any point $t$ . The derivative of $x$ with respect to $t$ is $\frac{dx}{dt} = \frac{d(t^3 + t)}{dt} = 3t^2 + 1$ . The derivative of $y$ with respect to $t$ is $\frac{dy}{dt} = \frac{d(t^4 + 2t^2)}{dt} = 4t^3 + 4t$ .
Evaluate Derivatives at $t=1$ : Now we need to evaluate these derivatives at $t=1$ to find the slope of the tangent line at that point. $\newline$ For $\frac{dx}{dt}$ , when $t=1$ , we have $\frac{dx}{dt} = 3(1)^2 + 1 = 3 + 1 = 4$ . $\newline$ For $\frac{dy}{dt}$ , when $t=1$ , we have $\frac{dy}{dt} = 4(1)^3 + 4(1) = 4 + 4 = 8$ .
Calculate Slope: The slope of the tangent line is given by $\frac{dy}{dx}$ , which is the ratio of $\frac{dy}{dt}$ to $\frac{dx}{dt}$ . At $t=1$ , the slope is $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{8}{4} = 2$ .
Find Coordinates at $t=1$ : Next, we need to find the coordinates of the point on the curve at $t=1$ to use in the point-slope form of the line equation. $\newline$ When $t=1$ , $x = 1^3 + 1 = 1 + 1 = 2$ , and $y = 1^4 + 2(1)^2 = 1 + 2 = 3$ . $\newline$ So the point on the curve at $t=1$ is $(2, 3)$ .
Use Point-Slope Form: Using the point-slope form of the line equation, $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is the point on the curve, we can write the equation of the tangent line. $\newline$ Substituting $m = 2$ and the point $(2, 3)$ , we get $y - 3 = 2(x - 2)$ .
Simplify Equation: Now we simplify the equation to get it into the form $y = mx + b$ . $\newline$ $y - 3 = 2x - 4$ $\newline$ Adding $3$ to both sides gives us $y = 2x - 1$ .