Find an equation for the line tangent to the curve at the point defined by the given value of t ...

Question:

Find an equation for the line tangent to the curve at the point defined by the given value of {eq}t{/eq}

• {eq}x=7\sin t, y=7 \cos t, t =\frac{3\pi}{4}{/eq}

A) {eq}y=7x+7\sqrt2{/eq} {eq}\enspace \enspace \enspace {/eq}B) {eq}y=-x+7\sqrt2{/eq} {eq}\enspace \enspace \enspace {/eq}C) {eq}y=x-7\sqrt2{/eq} {eq}\enspace \enspace \enspace {/eq}D) {eq}y=7\sqrt2x+1{/eq}

(Choose the correct one)

Tangent Lines to Parametric Curves

When you have a function {eq}y=f(x) {/eq} the equation of the tangent line at a point can be found by first finding the slope of the tangent line, given by {eq}\dfrac{dy}{dx} = f'(x) {/eq}. But what if the curve is defined parametrically as {eq}x=f(t) {/eq}, {eq}y=g(t) {/eq}? The slope of the tangent line is still given by the derivative of y with respect to x, and can be found by {eq}\dfrac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}} {/eq}

To find the equation of the tangent line, begin by finding the slope

{eq}\dfrac{dy}{dx} = \dfrac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \dfrac{dy}{dx} = \dfrac{-7\sin(t)}{7\cos(t)}\\ \dfrac{dy}{dx} = -\tan(t) {/eq}

And at {eq}t=\frac{3\pi}{4} {/eq} the slope is

{eq}-\tan\left(\frac{3\pi}{4}\right) = 1 {/eq}

We also need a point so that we can write the equation of the tangent line. When {eq}t=\frac{3\pi}{4} {/eq}, we have the point with coordinates {eq}x=7\sin\left(\frac{3\pi}{4}\right) = \frac{7\sqrt{2}}{2}\\ y=7\cos\left(frac{3\pi}{4}\right) = -\frac{7\sqrt{2}}{2} {/eq}

So the tangent line, in point slope form, is

{eq}y+\frac{7\sqrt{2}}{2} = 1\left(x-\frac{7\sqrt{2}}{2}\right) {/eq} which, in slope-intercept form is

{eq}y= x - 7\sqrt{2} {/eq}, answer choice C.