Home / AP Calculus AB 4.6 Approximating Values of a Function Using Local Linearity and Linearization- FRQs

AP Calculus AB 4.6 Approximating Values of a Function Using Local Linearity and Linearization- FRQs - Exam Style Questions

AP Calculus AB - FRQs and Answer - All Topics

No-Calc Question

Consider the curve \(G\) defined by the equation \(y^{3}-y^{2}-y+\tfrac{1}{4}x^{2}=0\).
A. Show that \(\dfrac{dy}{dx}=\dfrac{-x}{2\,(3y^{2}-2y-1)}\).
B. There is a point \(P\) on the curve \(G\) near \((2,-1)\) with \(x\)-coordinate \(1.6\). Use the line tangent to the curve at \((2,-1)\) to approximate the \(y\)-coordinate of point \(P\).
C. For \(x>0\) and \(y>0\), there is a point \(S\) on the curve \(G\) at which the line tangent to the curve at that point is vertical. Find the \(y\)-coordinate of point \(S\). Show the work that leads to your answer.
D. A particle moves along the curve \(H\) defined by the equation \(2xy+\ln y=8\). At the instant when the particle is at the point \((4,1)\), \(\dfrac{dx}{dt}=3\). Find \(\dfrac{dy}{dt}\) at that instant. Show the work that leads to your answer.

Most-appropriate topic codes (CED):

• TOPIC 3.2: Implicit Differentiation — parts (A) & (C) 
• TOPIC 4.6: Approximating Values of a Function Using Local Linearity and Linearization — part (B) 
• TOPIC 2.4: Connecting Differentiability and Continuity: When Derivatives Do and Do Not Exist (vertical tangent) — part (C) 
• TOPIC 4.5: Solving Related Rates Problems — part (D)
▶️ Answer/Explanation

(A) Show \(\dfrac{dy}{dx}=\dfrac{-x}{2(3y^{2}-2y-1)}\)
Start with \(y^{3}-y^{2}-y+\tfrac{1}{4}x^{2}=0\). Differentiate implicitly with respect to \(x\):
\(\dfrac{d}{dx}(y^{3})-\dfrac{d}{dx}(y^{2})-\dfrac{d}{dx}(y)+\dfrac{d}{dx}\!\left(\tfrac{1}{4}x^{2}\right)=0\)
\(3y^{2}\dfrac{dy}{dx}-2y\dfrac{dy}{dx}-\dfrac{dy}{dx}+\tfrac{1}{2}x=0\)
\((3y^{2}-2y-1)\dfrac{dy}{dx}+\dfrac{x}{2}=0\Rightarrow\dfrac{dy}{dx}=-\dfrac{x/2}{\,3y^{2}-2y-1\,}=\boxed{\dfrac{-x}{2(3y^{2}-2y-1)}}.\)

(B) Linearization at \((2,-1)\) and estimate at \(x=1.6\)
Slope at \((2,-1)\): \(m=\left.\dfrac{dy}{dx}\right|_{(2,-1)}=\dfrac{-2}{\,2\big(3(-1)^{2}-2(-1)-1\big)}=\dfrac{-2}{\,2(3+2-1)}=\dfrac{-2}{8}=-\tfrac{1}{4}.\)
Tangent line: \(y-(-1)=m(x-2)\Rightarrow y+1=-\tfrac{1}{4}(x-2).\)
For \(x=1.6\): \(y\approx -1-\tfrac{1}{4}(1.6-2)=-1-\tfrac{1}{4}(-0.4)=-1+0.1=\boxed{-0.9}.\)

(C) Vertical tangent for \(x>0,\ y>0\)
A vertical tangent occurs where \(\dfrac{dy}{dx}\) is undefined, i.e., denominator \(=0\) (and numerator \(\ne 0\)).
Set \(2(3y^{2}-2y-1)=0\Rightarrow 3y^{2}-2y-1=0\Rightarrow (3y+1)(y-1)=0.\)
Solutions \(y=-\tfrac{1}{3},\,1\). With \(y>0\), we get \(\boxed{y=1}\). (Since \(x>0\Rightarrow -x\ne 0\), the slope is indeed undefined/infinite.)

(D) Related rate \(\dfrac{dy}{dt}\) when \(x=4,\ y=1,\ \dfrac{dx}{dt}=3\)
Differentiate \(2xy+\ln y=8\) with respect to \(t\):
\(\dfrac{d}{dt}(2xy)+\dfrac{d}{dt}(\ln y)=0\Rightarrow 2\big(y\dfrac{dx}{dt}+x\dfrac{dy}{dt}\big)+\dfrac{1}{y}\dfrac{dy}{dt}=0.\)
Substitute \(x=4,\ y=1,\ \dfrac{dx}{dt}=3\): \(2\big(1\times 3+4\times \dfrac{dy}{dt}\big)+1\times \dfrac{dy}{dt}=0\).
\(6+8\dfrac{dy}{dt}+\dfrac{dy}{dt}=0\Rightarrow 9\dfrac{dy}{dt}=-6\Rightarrow \boxed{\dfrac{dy}{dt}=-\tfrac{2}{3}}.\)

More AP Calculus AB FRQ Questions..
Scroll to Top